JEE-Mathematics
Page No: 1
BINOMIAL THEOREM 1.
BINOMIAL EXPRESSION : Any algebraic expression which contains two dissimilar terms is called binomial expression. For example : x – y, xy +
2.
1 1 1 , 1, 3 etc. x z (x y )1 / 3
BINOMIAL THEOREM : The formula by which any positive integral power of a binomial expression can be expanded in the form of a series is known as BINOMIAL THEOREM. n
If x, y R and n N, then : ( x + y)n = nC0xn + nC1xn-1 y + nC2xn-2 y2 + ..... + nCrxn-r yr + ..... + nCnyn = n C r x n r y r r 0
This theorem can be proved by induction. Observations : (a) The number of terms in the expansion is ( n+1) i.e. one more than the index. (b) The sum of the indices of x & y in each term is n. (c) The binomial coefficients of the terms (nC0, nC1.....) equidistant from the beginning and the end are equal. i.e. n C r = n C r –1 (d)
n Symbol nCr can also be denoted by , C(n, r) or A nr . r
Some important expansions : (i)
(1 + x) n = n C 0 + n C 1 x + n C 2 x 2 + ........ + nC nx n .
(ii)
(1 – x) n = n C 0 – nC 1x + n C 2 x 2 + ........ + (–1) n . n C n x n .
Note :
The coefficient of x r in (1+x) n = nC r & that in (1–x) n = (–1) r . nC r
Illustration 1 :
Expand : (y + 2)6.
Solution :
6C y6 + 6C y5.2 + 6C y4.22 + 6C y3.23 + 6C y2. 24 + 6C y1 . 25 + 0 1 2 3 4 5
6C
6 6 . 2 .
= y6 + 12y5 + 60y4 + 160y3 + 240y2 + 192y + 64.
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Illustration 2 :
2y 2 Write first 4 terms of 1 5
Solution :
2y 2 7 2y 2 7 2y 2 7C , 7C C , C 3 , 2 0 1 5 5 5
Illustration 3 :
The value of
2
18 7 3.18.7.25 3
3
is -
3 6 6.243.2 15.81.4 20.27.8 15.9.16 6.3.32 64
(A) 1
Solution :
3
(B) 2
(C) 3
The numerator is of the form a 3 b 3 3ab a b a b Where, a = 18 and b = 7 Denominator can be written as
(D) 4 3
Nr = (18 + 7)3 = (25)3 6
3 6 6 C 1 .3 5 .21 6 C 2 .3 4 .2 2 6 C 3 .3 3 .2 3 6 C 4 3 2 .2 4 6 C 5 3 .2 5 6 C 6 2 6 3 2 5 6 25
Nr (25) 3 1 Dr (25) 3
3
Ans.
1
JEE-Mathematics Illustration 4 :
Page No: 2
If in the expansion of (1 + x)m (1 – x)n, the coefficients of x and x2 are 3 and – 6 respectively then m is -
[JEE 99]
(A) 6
Solution :
(B) 9
(C) 12
(D) 24
(m)(m 1).x 2 n(n 1) 2 ...... 1 nx x ...... (1 + x)m (1 – x)n = 1 mx 2 2 Coefficient of x = m – n = 3
........(i)
n(n 1) m (m 1) 6 2 2
Coefficient of x2 = –mn +
........(ii)
Solving (i) and (ii), we get m = 12 and n = 9. Do yourself - 1 : (i)
2 x Expand 3x 2
5
(ii) Expand (y + x)n 1
Pascal's triangle : A triangular arrangement of numbers as shown. The 1
numbers give the coefficients for the expansion of (x + y)n. The first row
1
is for n = 0, the second for n = 1, etc. Each row has 1 as its first and
1
last number. Other numbers are generated by adding the two numbers
IMPORTANT TERMS IN THE BINOMIAL EXPANSION (a)
3
1 3
1
1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 e tc.
immediately to the left and right in the row above.
3.
1 2
:
General term: The general term or the ( r +1)th term in the expansion of (x + y)n is given by Tr +1= nCr x n–r yr 11
2 1 The coefficient of x 7 in the expansion of ax bx
(b)
1 The coefficient of x –7 in the expansion of ax 2 bx
11
Also, find the relation between a and b, so that these coefficients are equal.
Solution :
(a)
2 1 In the expansion of ax bx
Tr + 1 =
11
2 11–r
C r(ax )
1 bx
11
, the general term is :
r
=
11
Cr.
a 11 r 22 3 r .x br
putting 22 – 3r = 7
3r = 15
T6 =
11
C5
r = 5
a6 7 .x b5
2 1 Hence the coefficient of x 7 in ax bx
11
is 11C 5a 6b –5.
Note that binomial coefficient of sixth term is 11C 5.
2
Ans.
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Illustration 5 : Find : (a)
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JEE-Mathematics
Page No: 3 (b)
1 In the expansion of ax 2 bx Tr + 1 =
11
C r(ax)
11–r
1 2 bx
11
, general term is :
r
= (–1) r 11 C r
a 11 r 11 3r .x br
putting 11 – 3r = –7
3r = 18
r = 6
T 7 = (–1) 6. 11 C 6
a 5 7 .x b6
1 Hence the coefficient of x –7 in ax 2 bx
11
is 11C 6a 5b –6.
Ans.
Also given : 2 1 Coefficient of x in ax bx
11
7
11
ab = 1
= coefficient of x
–7
1 in ax 2 bx
11
C 5 a 6 b –5 = 11 C 6 a 5 b –6 (
11
C 5 = 11 C 6 )
which is the required relation between a and b.
Ans.
Illustration 6 : Find the number of rational terms in the expansion of (9 1/4 + 8 1/6) 1000. The general term in the expansion of (9 1/4 + 8 1/6) 1000 is
Solution :
T r + 1 = 1000 C r
1000 r
9 1 4
r
8 = 1 6
1000
Cr
3
1000 r 2
r
22
The above term will be rational if exponents of 3 and 2 are integers It means
1000 r r and must be integers 2 2
The possible set of values of r is {0, 2, 4, ..........., 1000}
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Hence, number of rational terms is 501
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(b)
Ans.
Middle term : The middle term(s) in the expansion of (x + y)n is (are) : (i)
If n is even, there is only one middle term which is given by T(n +2)/2= nCn/2. x n/2. yn/2
(ii)
If n is odd, there are two middle terms which are T(n +1)/2 & T[(n+1)/2]+1
Important Note : Middle term has greatest binomial coefficient and if there are 2 middle terms their coefficients will be equal.
When r= n if n is even 2
nC r will be maximum
n–1 n+1 When r= or if n is odd 2 2
The term containing greatest binomial coefficient will be middle term in the expansion of (1 + x)n
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JEE-Mathematics
Page No: 4
x3 Illustration 7 : Find the middle term in the expansion of 3x 6 x3 The number of terms in the expansion of 3x 6
Solution :
9 1 i.e. 2
th
9 3 and 2
T 5 = T 4 +1
9
is 10 (even). So there are two middle terms.
th
are two middle terms. They are given by T 5 and T 6
x3 = C 4 (3x) 6 9
9
4
5
=
9
x 12 9.8.7.6 3 5 189 17 . x 17 = x = 4 1.2.3.4 2 4 .3 4 8 6
C 43 5x 5.
5
x 15 9.8.7.6 3 4 19 x3 21 19 . 5 5x = – x T 6 = T 5+1 = 9C 5(3x) 4 = – 9C 43 4.x 4. 5 = 6 1.2.3.4 2 .3 16 6
and (c)
Ans.
Term independent of x : Term independent of x does not contain x ; Hence find the value of r for which the exponent of x is zero. 10
Illustration 8 :
Solution :
x 3 2 The term independent of x in 2x 3 5 (A) 1 (B) 12 General term in the expansion is r
10 r
3r
is (C) 10 C1
(D) none of these
5 r
10 3 x2 3 2 3r 20 Cr 2 10 C r x 2 . 10 r 10 r For constant term, 3 2x 2 3 2 2 which is not an integer. Therefore, there will be no constant term. 10
Ans. (D)
Do yourself - 2 : 10
(i)
2 1 Find the 7 term of 3x 3
(ii)
2 3 Find the term independent of x in the expansion : 2x 3 x
th
25
7
(i i i )
Find the middle term in the expansion of :
(d)
Numerically greatest term : Let numerically greatest term in the expansion of (a + b) n be T r+1.
| Tr 1 | Tr Tr 1 Tr 2
where T r+1 = n C ra n–rb r
Solving above inequalities we get
Case I : When
(b)
2 1 2x x
n 1 n 1 1 r a a 1 1 b b
n 1 a is an integer equal to m, then Tm and Tm+1 will be numerically greatest term. 1 b
n 1 a is not an integer and its integral part is m, then T m+1 will be the numerically 1 b greatest term.
Case II : When
4
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6
2x 3 (a) 3 2x
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Page No: 5
JEE-Mathematics
Illustration 9 : Find numerically greatest term in the expansion of (3 – 5x) 11 when x =
Solution :
1 5
n 1 n 1 1 r a a Using 1 1 b b 11 1 11 1 1 r 3 3 1 1 5 x 5 x
solving we get 2 < r < 3 r = 2, 3 so, the greatest terms are T 2+1 and T 3+1 . Greatest term (when r = 2) T 3 = 11 C 2.3 9 (–5x) 2 = 55.3 9 = T 4 From above we say that the value of both greatest terms are equal. Ans. Illustration 10 : Given T 3 in the expansion of (1 – 3x) 6 has maximum numerical value. Find the range of 'x'. n 1 n 1 1 r Solution : Using a a 1 1 b b 6 1 7 1 2 1 1 1 1 3x 3x Let |x| = t 21t 21t 1 2 3t 1 3t 1
4t 1 1 1 21t 0 t , 3 3t 1 3t 1 3 4 15 t 2 0 t , 1 2 , 21t 2 3t 1 3 15 3t 1 Common solution
2 1 t , 15 4
2 2 1 1 x , , 15 15 4 4
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Do yourself -3 : (i) Find the numerically greatest term in the expansion of (3 – 2x)9, when x = 1.
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(ii)
4.
1 2x In the expansion of 2 3 possible integral values of n.
n
when x = –
1 , it is known that 3rd term is the greatest term. Find the 2
PROPERTIES OF BINOMIAL COEFFICIENTS : n
(1+x)n = C0 + C1x + C2x 2 + C3x 3 +.........+Cnx n n C r r r ; n N
....(i)
r 0
where C0,C 1,C 2,............Cn are called combinatorial (binomial) coefficients. (a) The sum of all the binomial coefficients is 2n. Put x = 1, in (i) we get n
C0 + C1 + C2 + .............+ Cn = 2n n C r 0
....(ii)
r 0
(b)
Put x=–1 in (i) we get n
C0 – C1 + C2–C3............+ Cn = 0 ( 1) r n C r 0 r 0
5
...(iii)
JEE-Mathematics (c)
Page No: 6
(d)
The sum of the binomial coefficients at odd position is equal to the sum of the binomial coefficients at even position and each is equal to 2n–1. From (ii) & (iii), C0 + C2 + C4 +............ = C1 + C3 + C5+....... = 2n–1 n n n+1 C r + C r–1= C r
(e)
n
n
Cr n r 1 r C r 1
(f)
n
Cr
n(n 1)(n 2).......(n r 1) n n 1 n n 1 n 2 C r 1 . C r 2 ....... r(r 1)(r 2)..........1 r r r 1
(g)
n
Cr
r 1 n 1 . C r 1 n 1
Illustration 11 : Prove that : 25C 10 + 24C 10 +........+10C 10 = 26C 11 Solution : LHS = 10C 10 + 11 C 10 + 12C 10 + ..............+ 25 C 10 11 C 11 + 11C 10 + 12C 10 + .......+ 25C 10 12 C 11 + 12 C 10 +........+ 25 C 10 13 C 11 + 13 C 10 +......... 25 C 10 and so on. LHS = 26C11 Aliter : LHS = coefficient of x10 in {(1 + x)10 + (1 + x)11 +............... (1+ x)25}
10
16 10 {1 x} 1 (1 x) in 1 x 1
coefficient of x
(1 x )26 (1 x )10 coefficient of x10 in x
coefficient of x11 in (1 x )
26
(1 x )10 = 26C11 – 0 = 26C11
Illustration 12 : Prove that : C1 + 2C 2 + 3C3 + ........ + nCn = n . 2n–1
(ii)
C0
(i)
n n n L.H.S. = r. n C r r. . n 1 C r 1 r 1 r 1 r
C1 C 2 C 2 n 1 1 ......... n 2 3 n 1 n 1
n
= n r 1
n 1
C r 1 n. n 1 C 0 n 1 C 1 ....... n 1 C n 1
= n . 2n–1 Aliter : (Using method of differentiation) (1 + x) n = nC0 + nC1x + nC2x2 + ....... + nCnxn
..........(A)
Differentiating (A), we get n(1 + x)n – 1 = C1 + 2C2x + 3C3x2 + ....... + n.Cnxn – 1. Put x = 1,
C 1 2C 2 3C 3 ........ n.C n n.2 n 1 n
(ii)
Cr 1 n n 1 n Cr L.H.S. = n 1 r 0 r 1 r 0 r 1 1 n n 1 1 n 1 C 1 n 1 C 2 ....... n 1 C n 1 = 1 2 n 1 1 C r 1 = n 1 n 1 r 0 n 1
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Solution :
(i)
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JEE-Mathematics
Page No: 7 Aliter : (Using method of integration) Integrating (A), we get
C x2 C x3 C x n 1 (1 x) n 1 C C 0 x 1 2 ........ n n 1 2 3 n 1 1 Put x = 0, we get, C = – n 1 n 1 C x2 C x3 C x n 1 (1 x) 1 C 0 x 1 2 ........ n n 1 2 3 n 1 Put x = 1, we get
(where C is a constant)
C1 C 2 C 2 n 1 1 ....... n 2 3 n 1 n 1 Put x = –1, we get C C 1 C 0 1 2 ....... 2 3 n 1 C0
n
Illustration 13 : If (1 + x)n = n C r x r , then prove that C 12 2.C 22 3.C 23 ......... n.C 2n r 0
Solution :
(1 + x)n = C0 + C1x + C2x2 + C2x3 + ........ + Cn xn
(2n 1)! ((n 1)!)2
.........(i)
Differentiating both the sides, w.r.t. x, we get n(1 + x)n–1 = C1 + 2C2x + 3C2x2 + ......... + n.Cnxn –1
.........(ii)
also, we have (x + 1)n = C0xn + C1xn – 1 + C2xn –2 + ......... + Cn
.........(iii)
Multiplying (ii) & (iii), we get (C1 + 2C2x + 3C3x2 + ........ + Cnxn – 1)(C0xn + C1xn –1 + C2xn – 2 + ......... + Cn) = n(1 + x)2n – 1 Equating the coefficients of xn – 1, we get
C 12 2C 22 3C 23 ......... n.C 2n n . 2n 1 C n 1 =
(2n 1)! ((n 1)!)2
Ans.
Illustration 14 : Prove that : C 0 – 3C 1 + 5C 2 – ........(–1) n(2n + 1)C n = 0 Solution : T r = (–1) r(2r + 1) nC r = 2(–1) rr . nC r + (–1) r nC r
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n n n n n T r = 2 ( 1) r .r. .n 1 C r 1 ( 1) r n C r = 2 ( 1) r .n 1 C r 1 ( 1) r . n C r r r 1 r 0 r 1 r 0
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= 2 n 1 C 0 n 1 C 1 ..... n C 0 n C 1 ....... = 0
Illustration 15 : Prove that ( 2nC 0) 2 – ( 2nC 1) 2 + ( 2nC 2) 2 – .... + (–1) n ( 2nC 2n) 2 = (–1) n. 2nC n Solution : (1 – x) 2n = 2n C 0 – 2n C 1x + 2nC 2 x 2 – ....+(–1) n 2n C 2nx 2n
...(i) and (x + 1) 2n = 2n C 0 x 2n + 2n C 1 x 2n–1 + 2n C 2 x 2n–2 +...+ 2n C 2n ....(ii) Multiplying (i) and (ii), we get (x 2 –1) 2n = ( 2n C 0 – 2nC 1x + .... + (–1) n 2n C 2nx 2n) × ( 2n C 0x 2n + 2nC 1 x 2n–1 + .... + 2n C 2n ) .... (iii) Now, coefficient of x 2n in R.H.S. = ( 2nC 0) 2 – ( 2nC 1) 2 + ( 2nC 2) 2 – ...... + (–1) n ( 2nC 2n) 2 General term in L.H.S., T r+1 = 2nC r(x 2) 2n – r(–1) r Putting 2(2n – r) = 2n r = n T n+1 = 2n C nx 2n (–1) n Hence coeffiecient of x 2n in L.H.S. = (–1) n. 2nC n But (iii) is an identity, therefore coefficient of x 2n in R.H.S. = coefficient of x 2n in L.H.S. ( 2nC 0) 2 – ( 2nC 1) 2 + ( 2nC 2) 2 – .... + (–1) n ( 2nC 2n) 2 = (–1) n. 2nC n
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JEE-Mathematics
Page No: 8
Illustration 16 : Prove that : nC 0. 2nC n – nC 1. 2n–2Cn n + nC 2. 2n–4Cn n + .... = 2 n Solution : L.H.S. = Coefficient of x n in [ nC 0(1 + x) 2n – nC 1(1 + x) 2n – 2 ......] = Coefficient of x n in [(1 + x) 2 – 1] n = Coefficient of x n in x n(x + 2) n = 2 n Illustration 17 : If (1 + x) n = C 0 + C 1 x + C 2 x 2 + ..... + C n x n then show that the sum of the products of the Ci's taken two at a time represented by :
C i C j is equal to 2
2n – 1
2n ! 2.n !n !
–
0 i j n
Solution :
Since (C 0 + C 1 + C 2 +.....+ C n–1 + C n) =
2
C 20 C 12 C 22 ..... C 2n 1 C 2n 2(C 0 C 1 C 0 C 2 C 0 C 3 ... C 0 C n + C 1 C 2 + C 1 C 3 +... + C 1 C n+ C 2 C 3 + C 2 C 4+...+C 2 C n +.....+C n–1 C n ) n 2
(2 ) = 2n C n + 2 C i C j 0 i j n
2n 1 Hence C i C j 2 0 i j n
2n ! 2.n !n !
Ans. 2
Illustration 18 : If (1 + x) n = C 0 + C 1x + C 2 x 2 +.....+ C nx n then prove that C i C j = (n – 1) 2nC n + 2 2n 0 i j n
Solution :
C i C j
L.H.S
2
0 i j n
= (C 0 + C 1 ) 2 + (C 0 + C 2 ) 2 +....+ (C 0 + C n ) 2 + (C 1 + C 2 ) 2 + (C 1 + C 3 ) 2 +....+ (C 1 + C n ) 2 + (C 2 + C 3 ) 2 + (C 2 + C 4 ) 2 +... + (C 2 + C n) 2 +....+ (C n – 1 + C n) 2 2
2
2
2
= n(C 0 C 1 C 2 .... C n ) 2 C i C j 0 i j n
2n ! 2 n 1 = n. 2n C n + 2. 2 {from Illustration 17} 2.n !n ! = n . 2n C n + 2 2n – 2n C n = (n – 1) . 2nC n + 2 2n = R.H.S. Do yourself - 4 : n n n n ........ = 0 1 2 n (A) 2 n – 1 (B) 2nCn
(i)
(D) 2 n+1
(a)
3C0 – 8C1 + 13C2 – 18C3 + .......... upto (n + 1) terms = 0, if n 2.
(b)
2C 0 + 2 2
(c)
C 20
C C1 C C 3 n 1 1 2 3 2 2 4 3 ...... 2 n 1 n 2 3 4 n 1 n 1
C 12 C 22 C2 (2n 1)! ...... n 2 3 n 1 ((n 1)!)2
MULTINOMIAL THEOREM : n
n n r r Using binomial theorem, we have (x + a)n = C r x a , n N r 0 n
n! x n r a r = n ! x s a r , where s + r = n = (n r)!r ! r 0 r s n r !s! This result can be generalized in the following form. (x1 + x2 + ...... + xk)n =
n! x1r1 x 2r2 ......x rkk r1 r2 ..... rk n r1 !r2 !.....rk !
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If (1 + x)n = C0 + C1x + C2x2 + .......... + Cnxn, n N. Prove that
(ii)
5.
(C) 2n
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JEE-Mathematics
Page No: 9 n! r1 r2 r3 rk The general term in the above expansion r !r !r !.....r ! . x 1 x 2 x 3 ......x k 1 2 3 k
The number of terms in the above expansion is equal to the number of non-negative integral solution of the equation r1 + r2 + ....... + rk = n because each solution of this equation gives a term in the above expansion. The number of such solutions is n k 1 C k 1 Particular cases : (i)
(x + y + z)n =
n!
r !s! t! x r y s z t
r s t n
n + 3 –1
The above expansion has (ii)
(x + y + z + u)n =
C3 – 1 = n + 2C2 terms
n! xp y q zr us p ! q !r !s! p q r s n
There are n+4–1C4–1 = n+3C3
terms in the above expansion.
Illustration 19 : Find the coefficient of x2 y3z4w in the expansion of (x – y – z + w)10 Solution :
(x – y – z + w)10 =
n! (x) p ( y) q ( z) r (w ) s p q r s 10 p ! q !r !s!
We want to get x2y3z4w this implies that p = 2, q = 3, r = 4, s = 1
Coefficient of x2y3z 4w is
10! (–1)3(–1)4 = –12600 2! 3! 4 ! 1!
Ans.
Illustration 20 : Find the total number of terms in the expansion of (1 + x + y)10 and coefficient of x2y3. Solution :
Total number of terms = 10+3–1C 3 – 1 = 12C2 = 66 Coefficient of x2y3 =
10! 2520 2! 3! 5 !
Ans.
Illustration 21 : Find the coefficient of x5 in the expansion of (2 – x + 3x2)6.
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Solution :
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The general term in the expansion of (2 – x + 3x2)6 =
6! r 2 ( x) s (3x 2 ) t , where r + s + t = 6. r !s ! t !
6! r 2 ( 1) s (3) t x s 2 t r !s ! t !
=
For the coefficient of x5, we must have s + 2t = 5. But, r + s + t = 6,
s = 5 – 2t and r = 1 + t, where 0 r, s, t 6.
Now t = 0
r = 1, s = 5.
t = 1
r = 2, s = 3.
t=2
r = 3, s = 1.
Thus, there are three terms containing x5 and coefficient of x5 =
6! 6! 6! 21 ( 1)5 3 0 2 2 ( 1)3 31 2 3 ( 1)1 3 2 1! 5 ! 0! 2 ! 3 ! 1! 3! 1! 2 !
= –12 – 720 – 4320 = –5052.
Ans.
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JEE-Mathematics
Page No: 10 2n
n 1
r
Illustration 22 : If (1+x+x 2)n a r x , then prove that
(a) ar = a2n–r
r 0
Solution :
(a)
(b) a r r 0
1 n (3 a n ) 2
We have 2n
n
1 x x a x 2
r
....(A)
r
r 0
Replace x by
1 x n
2n 1 1 1 1 x 2 a r x x r 0
x x 1 a x
n
2
2n
r
2n r
r
r 0
2n
2n
r 0
r 0
a r x r a r x 2n r
(b)
{Using (A)}
Equating the coefficient of x2n–r on both sides, we get a2n–r = ar for 0 < r < 2n. Hence ar = a 2n–r. Putting x=1 in given series, then a 0 + a 1 + a 2 + .........+ a 2n = (1+1+1) n a0 + a 1 + a 2 + ..........+ a 2n = 3 n ....(1) But ar = a2n–r for 0 < r < 2n series (1) reduces to 2(a0 + a 1 +a 2 + ........+ a n–1) + a n = 3 n.
a0 + a 1 +a2 + .......+ an–1 =
1 n (3 – an) 2
Do yourself - 5 : (i)
APPLICATION OF BINOMIAL THEOREM :
Illustration 23 : If 6 6 14
2 n 1
= [N] + F and F = N – [N]; where [.] denotes greatest integer function, then
NF is equal to (A) 20 2n+1
Solution :
Since 6 6 14
(B) an even integer 2 n 1
(D) 40 2n+1
= [N] + F
Let us assume that f = 6 6 14 Now, [N] + F – f
(C) odd integer
= 6 6 14
2 n 1
2 n 1
; where 0 f < 1.
– 6 6 14
2n 1
2n 2 n 2 = 2 2 n 1 C 1 6 6 (14) 2 n 1 C 3 6 6 (14) 3 ....
[N] + F – f = even integer. Now 0 < F < 1 and 0 < f < 1 so –1 < F – f < 1 and F – f is an integer so it can only be zero Thus NF = 6 6 14
2 n 1
6 6 14 2n 1 = 20 10
2n+1
.
Ans. (A,B)
NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#11\Eng\01.BINOMIAL.p65
6.
Find the coefficient of x2y5 in the expansion of (3 + 2x – y)10.
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Page No: 11
JEE-Mathematics
Illustration 24 : Find the last three digits in 1150. Solution :
Expansion of (10 + 1) 50 = 50C 010 50 + 50C 110 49 + ..... + 50C 4810 2 + 50C 4910 + 50C 50 = 50 C 0 10 50 50 C 1 10 49 ...... 50 C 47 10 3 + 49 × 25 × 100 + 500 + 1 1000 K
1000 K + 123001
Last 3 digits are 001.
Illustration 25 : Prove that 22225555 + 55552222 is divisible by 7. When 2222 is divided by 7 it leaves a remainder 3. So adding & subtracting 35555, we get :
Solution :
5555 5555 E 2222 3 5555 5555 2222 3 E1
E2
For E1 : Now since 2222–3 = 2219 is divisible by 7, therefore E1 is divisible by 7 ( xn – an is divisible by x –a) For E2 : 5555 when devided by 7 leaves remainder 4. So adding and subtracting 42222, we get : E2 = 35555 + 42222 + 55552222 – 42222 = (243)1111 + (16)1111 + (5555)2222 – 42222 Again (243)1111 + 16 1111 and (5555) 2222 – 4 2222 are divisible by 7 ( xn + an is divisible by x + a when n is odd) Hence 22225555 + 55552222 is divisible by 7. Do yourself - 6 : (i)
Prove that 525 – 325 is divisible by 2.
(ii)
Find the remainder when the number 9100 is divided by 8.
(i i i )
Find last three digits in 19100.
(iv)
Let R (8 3 7 )20 and [.] denotes greatest integer function, then prove that : (a) [R] is odd
(v)
NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#11\Eng\01.BINOMIAL.p65
7.
E
(b)
R [R ] 1
1 (8 3 7 )20
Find the digit at unit's place in the number 171995 + 111995 – 71995.
BINOMIAL THEOREM FOR NEGATIVE OR FR ACTIONAL INDICES : If n Q, then (1 + x)n = 1 + nx +
n(n 1) 2 n(n 1)(n 2) 3 x + x + ....... provided | x | < 1. 2! 3!
Note : (i)
When the index n is a positive integer the number of terms in the expansion of ( 1+ x)n is finite i.e. (n+1) & the coefficient of successive terms are : nC0, nC1, nC2, ....... nCn
(ii)
When the index is other than a positive integer such as negative integer or fraction, the number of terms in the expansion of (1+ x)n is infinite and the symbol nCr cannot be used to denote the coefficient of the general term.
(iii)
Following expansion should be remembered (|x| < 1). (a)
(1 + x)-1 =1 – x + x2 – x3 + x4 - ....
(b)
(1 – x)–1 =1 + x + x2 + x3 + x4 + ....
(c)
(1 + x)-2 =1 – 2x + 3x2 – 4x3 + ....
(d)
(1 – x)–2 =1 + 2x + 3x2 + 4x3 + ....
(e)
(1 + x)–3 = 1 – 3x + 6x2 – 10x3 + ..... +
11
( 1) r (r 1)(r 2) r x ........ 2!
JEE-Mathematics
(r 1)(r 2) r x ........ 2! The expansions in ascending powers of x are only valid if x is ‘small’. If x is large i.e. | x |>1 then we
(f) (iv)
Page No: 12
(1 – x)–3 = 1 + 3x + 6x2 + 10x3 + ..... +
may find it convenient to expand in powers of 1/x, which then will be small. 8.
APPROXIM ATIONS : (1 + x)n = 1 + nx +
n(n 1) 2 n(n 1)(n 2) 3 x + x ....... 1.2.3 1.2
If x < 1, the terms of the above expansion go on decreasing and if x be very small, a stage may be reached when we may neglect the terms containing higher powers of x in the expansion. Thus, if x be so small that its square and higher powers may be neglected then (1 + x)n = 1 + nx, approximately. This is an approximate value of (1 + x)n
Illustration 26 : If x is so small such that its square and higher powers may be neglected then find the approximate
Solution :
(1 3x )1 / 2 (1 x) 5 / 3
4 x 1 / 2
(1 3x )1 / 2 (1 x) 5 / 3
4 x 1 / 2
3 5x x 1 1 / 2 1 19 x 1 19 x 2 3 x 1 2 x 1 = = 2 1/2 2 6 8 2 6 4 x 2 1 4
1
=
1 x 19 x 19 41 x = 1 x = 1 – x 2 2 4 6 8 12 24 Illustration 27 : The value of cube root of 1001 upto five decimal places is – =
(A) 10.03333
Solution :
(B) 10.00333
(1001) 1/3 = (1000+1) 1/3 = 10 1
(C) 10.00033 1 1000
1/3
Ans.
(D) none of these
1 1 1 / 3(1 / 3 1) 1 ..... = 10 1 . 3 1000 2! 1000 2
= 10{1 + 0.0003333 – 0.00000011 + .....} = 10.00333
Illustration 28 : The sum of 1 +
(A)
Solution :
1 1.3 1.3.5 .... is 4 4.8 4.8.12 (B)
2
Ans. (B)
1
(C)
2
Comparing with 1 + nx +
(D) 2 3/2
3
n(n 1) 2 x .... 2!
nx = 1/4
....... (i) 2
and
n(n 1)x 1.3 = 2! 4.8
or
1 1 3 nx(nx x) 3 = x 4 4 16 2! 32
1 3 1 1 3 x x 4 4 4 4 2
(by (i))
........(ii)
putting the value of x in (i) n (–1/2) = 1/4 n = –1/2
sum of series = (1 + x) n = (1 – 1/2) –1/2 = (1/2) –1/2 =
12
2
Ans. (A)
NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#11\Eng\01.BINOMIAL.p65
value of
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JEE-Mathematics
Page No: 13 9.
EXPONENTIAL SERIES : (a) e is an irrational number lying between 2.7 & 2.8. Its value correct upto 10 places of decimal is 2.7182818284. (b)
Logarithms to the base ‘e’ are known as the Napierian system, so named after Napier, their inventor. They are also called Natural Logarithm.
10.
(c)
1 x x2 x 3 ....... ; where x may be any real or complex number & e = Lim 1 e = 1 n n 1! 2 ! 3 !
(d)
ax = 1
(e)
e = 1
x
x x2 2 x3 3 na n a n a ....... , where a > 0 1! 2! 3!
1 1 1 ....... 1! 2 ! 3 !
LOGARITHMIC SERIES : (a)
n (1 + x) = x
x2 x3 x4 ....... , where –1 < x 1 2 3 4
(b)
n (1 - x) = - x
x2 x 3 x 4 ....... , where –1 x 1 2 3 4
NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#11\Eng\01.BINOMIAL.p65
Remember :
E
1 1 1 ....... n 2 2 3 4
(i)
1
(iii)
n2 = 0.693
(ii)
elnx = x ; for all x > 0
(iv)
n10 = 2.303
ANSWERS FOR DO YOURSELF 2
1.
(i)
5
x x C 0x (3x ) + C 1(3x ) + 5C 2(3x 2) 3 2 2 2 5
5
2 4
3
2 2
(ii) nC0y n + nC 1y n–1.x + nC 2.y n–2.x 2 + ..........+ nCn.x n 70 8 x ; 3
(ii)
25 ! 15 10 2 3 ; 10! 5 !
(i i i ) (a) –20; (b) –560x5, 280x2
2 :
(i)
3. 4. 5. 6.
(i) 4th & 5th i.e. 489888 ( i i ) n = 4, 5, 6 (i) C (i) –272160 or – 10C5 × 5C2 × 108 (ii) 1 (i i i ) 8 0 1 (v) 1
13
4
x x x + C 3(3x ) + 5C 4(3x 2) 1 + 5C 5 2 2 2 5
5
n
JEE-Mathematics
Page No: 14
EXERCISE - 01
CHECK YOUR GRASP
SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) n
1.
x If the coefficients of x7 & x8 in the expansion of 2 are equal , then the value of n is 3 (A) 15 (B) 45 (C) 55 (D) 56 n
2.
1 The sum of the binomial coefficients of 2 x is equal to 256 . The constant term in the expansion x is -
(A) 1120 3.
(B) 2110
(C) 1210
(D) none
The sum of the coefficients in the expansion of (1 2x + 5x2)n is ' a ' and the sum of the coefficients in the expansion of (1 + x)2n is b . Then (B) a = b2
(A) a = b 4.
(C) a2 = b
(D) ab = 1
Given that the term of the expansion (x1/3 x1/2)15 which does not contain x is 5 m where m N , then m is equal to (A) 1100
5.
(B) 1010
(D) none
1 4 x 1 7 1 4 x 1 7 is a polynomial in x of degree 2 2 4 x 1 1
The expression (A) 7
6.
(C) 1001
(B) 5 1/3
In the binomial (2
(C) 4
(D) 3
1/3 n
+3
) , if the ratio of the seventh term from the beginning of the expansion to the
seventh term from its end is 1/6 , then n is equal to (A) 6
(B) 9
(C) 12
(D) 15
11
7.
3 2 The term independent of x in the product (4 x 7x ) x x
is -
(A) 7. 11 C 6 8.
(B) 36. 11C6 (C) 35. 11C5 (D) –12. 211 If ‘a’ be the sum of the odd terms & ‘b’ be the sum of the even terms in the expansion of (1 + x)n , then (1 x²)n is equal to (A) a² b²
(D) none
The sum of the coefficients of all the even powers of x in the expansion of (2x2 3x + 1)11 is (A) 2 . 610
10.
(C) b² a²
(B) 3 . 610
(C) 611
(D) none
The greatest terms of the expansion (2x + 5y)13 when x = 10, y = 2 is (A)
13
C5 . 208 . 105
(B)
13
C6 . 207 . 104
(C)
13
C4 . 209 . 104
(D) none of these
100
11.
Number of rational terms in the expansion of (A) 25
12.
(B) 26
is -
(C) 27
(D) 28
(C) 22n–1
(D) 2nCn
(C) 13
(D) 14
n p If 0 for p < q, where p, q W, then r 0 2r q
(A) 2n 13.
2 4 3
(B) 2n–1
x 47 5 52 j x , then y = 4 j 1 3 y (A) 11
(B) 12
14
NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#11\Eng\01.BINOMIAL.p65
9.
(B) a² + b²
E
JEE-Mathematics
Page No: 15 14.
If n N & n is even , then (A) 2n
15.
(B)
17.
(B) 3 31
(D) none of these
(C) 262
(D) 1
(C) 25C10
(D) 25C11
C0 C C C 1 2 ...... 10 is equal to (here Cr = 10Cr) 1 2 3 11 211 11
(B) n
If a n n r 0
211 1 11
(C)
311 11
(D)
311 1 11
n 1 r , then equals n Cr r 0 C r
(A) ( n -1) an 19.
(C) 2n n !
10 10 15 The value of is equal to r 0 r 14 r (A) 25C12 (B) 25C15
(A)
18.
2 n 1 n!
Let R (5 5 11) 31 I ƒ , where I is an integer and ƒ is the fractional part of R, then R · ƒ is equal to (A) 231
16.
1 1 1 1 ...... = 1. (n 1) ! 3 ! (n 3) ! 5 ! (n 5) ! (n 1) ! 1 !
[JEE 98]
(B) n an 400
The last two digits of the number 3 (A) 81
(C) n an /2
(D) none of these
(C) 29
(D) 01
are -
(B) 43
SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 20.
21.
22.
If the coefficients of three consecutive terms in the expansion of (1 + x)n are in the ratio of 1 : 7 : 42, then n is divisible by (A) 9 (B) 5 (C) 3 (D) 11 1 In the expansion of 3 4 4 6
20
-
(A) the number of irrational terms = 19
(B) middle term is irrational
(C) the number of rational terms = 2
(D) 9th term is rational
If (1 + x + x 2 + x 3)100 = a 0 + a1x + a 2x 2 + ......... + a 300x300, then -
NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#11\Eng\01.BINOMIAL.p65
(A) a0 + a1 + a2 + a3 +.......+ a300 is divisible by 1024
E
(B) a0 + a 2 + a 4 +.......+ a 300 = a 1 + a 3 + ....... + a 299 (C) coefficients equidistant from beginning and end are equal (D) a1 = 100 23.
The number 101100 1 is divisible by (A) 100
(B) 1000
(C) 10000
(D) 100000
n
24.
25.
If
9 80 = I + f where I , n are integers and 0 < f < 1 , then -
(A) I is an odd integer
(B) I is an even integer
(C) (I + f) (1 f) = 1
(D) 1 f =
1 In the expansion of x 2 / 3 x
9 80
n
30
, a term containing the power x13 -
(A) does not exist
(B) exists and the co-efficient is divisible by 29
(C) exists and the co-efficient is divisible by 63
(D) exists and the co-efficient is divisible by 65
15
JEE-Mathematics
The co-efficient of the middle term in the expansion of (1 + x)2n is (A)
1.3.5 .7......(2 n 1) 2n n!
(B)
2n
(C)
(n 1) (n 2) (n 3) .... (2n 1) (2n) 1.2 .3.......... (n 1) n
(D)
2 .6 .10 .14 ...... (4n 6) (4n 2) 1.2.3 .4 .....(n 1) . n
CHE CK Y OU R G R ASP
Cn
ANSWER
KEY
EXERCISE-1
Que.
1
2
3
4
5
6
7
8
9
10
Ans.
C
A
A
C
D
B
B
A
B
C
Que.
11
12
13
14
15
16
17
18
19
20
B
C
D
B,D
Ans.
B
B
C
B
C
D
Que.
21
22
23
24
25
26
A, B, C
A ,C , D
B,C,D
A,B, C,D
Ans.
A,B, C,D A,B, C,D
16
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26.
Page No: 16
E
JEE-Mathematics
Page No: 17
EXERCISE - 02
BRAIN TEASERS
SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 1.
The coefficient of xr (0 r n 1) in the expression : (x + 2)n1 + (x + 2)n2 . (x + 1) + (x + 2)n3 . (x + 1)² + ...... + (x + 1)n1 is (A)
2.
n
Cr (2r 1)
(B)
n
Cr (2nr 1)
(C)
(D)
n
Cr (2nr + 1)
(B) odd & of the form 3n
(C) odd & of the form (3n 1)
(D) odd & of the form (3n + 1)
4
2 12
The co-efficient of x in the expansion of (1 x + 2x ) (A)
4.
Cr (2r + 1)
If (1 + x + x²)25 = a0 + a1x + a2x² + ..... + a50 . x50 then a0 + a2 + a4 + ..... + a50 is (A) even
3.
n
12
C3
(B)
13
C3
is (C)
14
(D) 12C3 + 3 13C 3 + 14C 4
C4
Let (1 + x2)2 (1 + x)n = A0 + A1 x + A2 x2 + ...... If A0, A1, A2 are in A.P. then the value of n is (A) 2
(B) 3
(C) 5
(D) 7
nr
5.
If
nk
Cr = xCy then -
k 1
(A) x = n + 1 ; y = r
(B) x = n ; y = r + 1
(C) x = n ; y = r 6.
(D) x = n + 1 ; y = r + 1
t
Co-efficient of in the expansion of ( + p)
m1
+ ( + p)
m2
( + q) + ( + p)m 3 ( + q)2 +...... ( + q)m 1 where
q and p q is m
(A) 7.
Ct pt qt
m
(B)
p q
NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#11\Eng\01.BINOMIAL.p65
p q
Ct pt qt
m
(D)
p q
C t pm t q m t pq
(C) 2
(D) 2
(B) 5
The value r for which
(C) 6
30r 15r 30r 1 151 ....... 300 15r is maximum is/are (B) 22
th
(D) 8
(C) 23
3 x term in the expansion of 2 3
(D) 24
n
10.
If the 6
11.
value(s) of n can be (A) 11 (B) 12 (C) 13 (D) 14 In the expansion of (1 + x)n (1 + y)n (1 + z)n , the sum of the co-efficients of the terms of degree ' r ' is (C)
r
3n
Cr
(D) 3 . 2nCr
35 10 45 r x , then x – y is equal to 6 r 0 5 y (A) 39
(B) 29 s
13.
when x = 3 is numerically greatest then the possible integral
(B) n C 3
3 (A) n C r
12.
Number of terms free from radical sign in the expansion of (1 + 31/3 + 71/7)10 is -
(A) 21
E
m
(C)
(B) 1
(A) 4 9.
The co-efficient of x401 in the expansion of (1 + x + x2 + ...... + x9) 1 , (x < 1) is (A) 1
8.
C t pm t q m t
The value of
(C) 52
(D) 40
(C) 3 n
(D) 3(3 n – 1)
n
n C s s C r is r 0 s 1 r s
(A) 3 n – 1
(B) 3 n + 1
17
JEE-Mathematics 14.
Page No: 18
log 2 3 In the expansion of x 3 .2
x3
11
-
(A) there appears a term with the power x2 (B) there does not appear a term with the power x2 (C) there appears a term with the power x 3
1 3 The sum of the series (1² + 1).1! + (2² + 1).2! + (3² + 1).3! + ..... + (n² + 1).n! is (A) (n + 1) . (n + 2)! (B) n . (n + 1)! (C) (n + 1) .(n + 1)! (D) none of these (D) the ratio of the co-efficient of x3 to that of x 3 is
16.
1 The binomial expansion of x k 2 k x
3n
,
n N contains a term independent of x -
(A) only if k is an integer
(B) only if k is a natural number
(C) only if k is rational 17.
(D) for any real k
n
2
n
Let n N. If (1 + x) = a0 + a1x + a2x + ........+ anx and an–3, an–2, an–1 are in AP, then (A) a1, a2, a3 are in AP
(B) a1, a2, a3 are in HP
(C) n = 7 18.
(D) n = 14
Set of values of r for which, (A) 4 elements
18
Cr 2 + 2 .
18
Cr 1 +
18
Cr 20C13 contains -
(B) 5 elements
BRAIN TEASER S
(C) 7 elements
ANSWER
KEY
(D) 10 elements
EXERCISE-2
Que.
1
2
3
4
5
6
7
8
9
10
Ans.
B
A
D
A,B
B
B
B
C
B,C
B,C,D
Que.
11
12
13
14
15
16
17
18
Ans.
C
D
A
B,C,D
B
D
A,C
C
18
NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#11\Eng\01.BINOMIAL.p65
15.
E
JEE-Mathematics
Page No: 19
EXERCISE - 03
MISCELLANEOUS TYPE QUESTIONS
FILL IN THE BLANKS 1.
The greatest binomial coefficient in the expansion of (a + b)n is ________ given that the sum of all the coefficients is equal to 4096.
2.
The number 71995 when divided by 100 leaves the remainder ________. 15
3. 4. 5.
2 1 The term independent of x in the expansion of x is ________ . x p n np If (1 + x + x² + ..... + x ) = a0+ a1x + a2x² + ..... + anp x then a1+ 2 a2+ 3a3 + .... + npanp = ________ . If (1 + x) (1 + x + x2) (1 + x + x2 + x3) ...... (1 + x + x2 + x3 + ...... + xn) a0 + a1x + a2x2 + a3x3 + ...... + amxm m
then
a
r
has the value equal to ________ .
r 0 8
6.
1 2 If the 6th term in the expansion of the binomial 8 / 3 x log10 x x
7.
(1 + x) (1 + x + x2) (1 + x + x2 + x3) ...... (1 + x + x2 + ...... + x100) when written in the ascending power of
is 5600, then x = ________ .
x then the highest exponent of x is ________ . MATCH THE COLUMN Following question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given statement in Column-I can have correct matching with ONE statement in Column-II. 1.
Column-I
Column-II
(A)
(2n + 1) (2n + 3) (2n + 5) ....... (4n 1) is equal to
(p)
(n 1) n n!
(B)
C1 2 . C2 3 . C3 n . Cn is equal to ...... C0 C1 C2 C n 1
(q)
n . 2n . (2n 1)
(r)
(4n) ! n ! 2 . (2n) ! (2n) !
(s)
n (n 1) 2
here Cr stand for (C)
n
Cr .
If (C0 + C1) (C1 + C2) (C2 + C3) ...... (Cn1 + Cn)
n
= m . C1C2C3 .... Cn1 , then m is equal to (D)
If Cr are the binomial coefficients in the expansion of
NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#11\Eng\01.BINOMIAL.p65
n
E
n
(1 + x)n, the value of (i j) Ci Cj is i 1 j 1
ASSERTION
&
REASON
These questions contains, Statement I (assertion) and Statement II (reason). (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I (C) Statement-I is true, Statement-II is false (D) Statement-I is false, Statement-II is true 1.
Statement-I : Coefficient of ab 8 c 3 d 2 in the expansion of (a + b + c + d) 14 is 180180 Because Statement-II : General term in the expansion of (a 1 + a 2 + a 3 + ..... + a m ) n =
n!
n !n !n !....n ! a1n a 2n ...a nm , where n 1 + n 2 + n 3 + ... + n m = n. 1
(A) A
2
3
1
2
m
m
(B) B
(C) C
19
(D) D
JEE-Mathematics 2.
Page No: 20
Statement-I : If q =
1 and p + q = 1, then 3
15
r
15
C r p r q 15 r 15
r 0
1 5 3
Because n n
r
r C p q
Statement-II : If p + q = 1 , 0 < p < 1, then
n r
r
np
r 0
(A) A 3.
(B) B
(C) C
(D) D
Statement-I : The greatest value of 40 C 0 . 60 C r + 40 C 1 . 60 C r–1 ........ 40 C 40 . 60 C r–40 is 100 C 50 Because Statement-II : The greatest value of 2n C r , (where r is constant) occurs at r = n. (A) A
4.
(B) B
(C) C
(D) D
Statement-I : If x = n C n–1 + n+1 C n–1 + n+2 C n–1 + .......... + 2n C n–1 , then
x 1 is integer.. 2n 1
Because Statement-II : n C r + n C r–1 = n+1 C r and n C r is divisible by n if n and r are co-prime. (A) A
(B) B
(C) C
(D) D
COMPREHENSION BASED QUESTIONS Comprehension # 1 2n
If n is positive integer and if (1 + 4x + 4x 2) n =
a r x r , where ai's are (i = 0, 1, 2, 3, ..... , 2n) real r 0
numbers. On the basis of above information, answer the following questions : n
1.
The value of 2 a 2r is r 0
(A) 9 n – 1
(B) 9 n + 1
(C) 9 n – 2
(D) 9 n + 2
(B) 9 n + 1
(C) 9 n – 2
(D) 9 n + 2
(B) (n – 1).2 2n
(C) n.2 2n
(D) (n + 1).2 2n
(B) 8n 2 – 4
(C) 8n 2 – 4n
(D) 8n – 4
n
2.
The value of 2 a 2r 1 is r 1
(A) 9 n – 1 The value of a 2n–1 is (A) 2 2n 4.
The value of a 2 is (A) 8n
ANSWER
M ISCEL L AN E OU S TYP E Q U ESTION
2. 43
3 . 3003
4.
np (p + 1) n 2
5 . (n + 1) !
Matc h th e C o lu mn 1 . (A)(r), (B)(s), (C)(p), (D)(q)
A s s er ti o n & R eas o n 1. C
EXERCISE-3
F i ll i n t h e B lanks 1 . 12 C 6
KEY
2. D
C o mp r eh e ns i o n
3. C
B as ed
Comprehensi on # 1 :
4. A
Qu es ti o ns 1. B
2. A
3. C
20
4. C
6 . x = 10
7 . 5050
NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#11\Eng\01.BINOMIAL.p65
3.
E
JEE-Mathematics
Page No: 21
EXERCISE - 04 [A]
CONCEPTUAL SUBJECTIVE EXERCISE
1.
If the coefficients of (2r + 4)th, (r - 2)th terms in the expansion of (1 + x)18 are equal, find r.
2.
If the coefficients of the rth, (r +1)th & (r + 2 )th terms in the expansion of (1 + x)14 are in AP, find r.
3.
x 3 Find the term independent of x in the expansion of : (a) 2 3 2x
4.
Prove that :
5.
If
40
6.
If
n + 1
7.
Which is larger : (9950 +10050) or (101)50.
8.
Show that 2n -2Cn-2 + 2. 2n -2Cn-1+ 2n -2Cn >
9.
Find the coefficient of x4 in the expansion of :
10
(a) 10.
n -1
1 1 / 3 x 1 / 5 (b) x 2
8
Cr + n -2Cr+ n -3Cr +....... + rCr= n Cr+1.
C1 . x(1 x)39 + 2 . 40C2 x2 (1 x)38 + 3 40C3 x3 (1 x)37 + ...... + 40. 40C40 x40 = ax + b, then find a & b. C2 + 2 (2C2 + 3C2 + 4C2 + ...... + nC2) = 12 + 22 + 32 + ......... + 1002, then find n.
4n , n N, n >2 n 1
(1 + x + x2 + x3)11
(b)
(2 – x + 3x2)6
Find numerically the greatest term in the expansion of : (a)
( 2 + 3x)9 when x =
3 2
(b)
( 3 – 5x)15 when x =
1 5
11.
2 Prove that the ratio of the coefficient of x10 in (1– x2)10 & the term independent of x in x x
12.
3x 2 1 . Find the term independent of x in the expansion of (1+ x +2x ) 3x 2
13.
Prove that C k sin Kx. cos ( n - K)x = 2n -1 sin nx.
10
is 1 : 32.
9
3
n n
k 0
NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#11\Eng\01.BINOMIAL.p65
14.
E
Find the coefficient of : (a)
x6 in the expansion of (ax2 + bx + c)9.
(c)
a2 b3 c4d in the expansion of (a – b – c + d)10.
(b)
25
30 70 100 Prove that : C 25 r 0 r 25 r
20
20 30 r 25 r x C y , then find x, y. (b) r 0
15.
(a)
16.
n r n n k Prove that : r k k r k
17.
25 r 30 30 Prove that : 1 0 r 25 r r 0
18.
n 2 n 1 n 2n 1 Prove that : n 2 r 0 r r 2
x2 y3 z4 in the expansion of (ax - by + cz)9.
21
JEE-Mathematics
Page No: 22
Prove the following (here C r = n C r) (Q. 19 to 26) :
(2n)! (n 1)!(n 1)!
19.
C 0C 1 + C 1C 2 + C 2C 3 +.......+ C n-1C n =
20.
C 0 C r C 1 C r 1 C 2 C r 2 ....... C n r C n
21.
C02 + C 12 + C 22 +....... + C n2 =
22.
C 0 C 1 C 2 C 3 ....... ( 1) r .C r
23.
C1 + 2C2+ 3C3 + ....... + n. Cn =n. 2n-1
24.
C0 + 2C1+ 3C2 +....... + (n +1) Cn= (n + 2) 2n-1
25.
C0
26.
C 1 2C 2 3C 3 n.C n n(n 1) ...... C0 C1 C2 C n 1 2
27.
Prove the identity
28.
If (1 + x )15 = C0 + C1. x + C2. x2 +....+ C15. x15 and C2 + 2C3 + 3C4 +....+ 14C15 = a2 + c, then find a + b + c.
29.
Evaluate : 2
2n ! (n r)!(n r)!
(2n)! n !n ! ( 1) r (n 1)! r !.(n r 1)!
C1 C 2 C 2 n 1 1 ....... n 2 3 n 1 n 1
1 2 n 1
1 2n 2 1 2 n 1 . Cr C r 1 2n 1 2 n C r b
2 0 15
14 30 29
2 1 14
CON CEP TUAL SU BJ ECTIVE E X ER CISE
1. r = 6
2.
r = 5 or 9
7 . 101 5 0
9.
(a) 990
3.
x = 50, y = 25
30 15 ...... 2 13 15 0
ANSWER (a) T3 =
5 12
28. 28
KEY
(b) T6 = 7
EXERCISE-4(A)
5. a = 40, b = 0
17 7.3 13 (b) 455 x 312 12. 54 2 (b) –1260.a2b 3c4 ; (c) –12600
(b) 3660 1 0 . (a) T7
1 4 . (a) 84b 6c 3 + 630ab 4c 4 +756a 2 b 2c 5 + 84a 3c 6 ; 1 5 (a)
13 30 28
30 29. 15
22
6. 100
NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#11\Eng\01.BINOMIAL.p65
15 30 30
E
JEE-Mathematics
Page No: 23
EXERCISE - 04 [B] 2n
1.
3.
SUBJECTIVE EXERCISE
2n
r r If a r (x 2) b r (x 3) & ak =1 for all k n, then show that bn = 2n+1Cn+1 r 0
2.
BRAIN STORMING
r 0
Prove the following : (a)
if n is odd 0 2 2 2 2 2 C 0 C 1 C 2 C 3 ....... ( 1) n C n n/2 n C n / 2 if n is even ( 1)
(b)
1.C 02 + 3.C 12 + 5.C 22 +.......+(2n+1)C n2 =
(a)
x 2 Find the index n of the binomial 5 5
(n 1)(2n)! n !n !
n
if the 9th term of the expansion has numerically the greatest
coefficient (n N). (b) 4.
For which positive values of x is the fourth term in the expansion of (5 +3x)10 is the greatest.
If a0, a1, a2, ....... be the coefficients in the expansion of (1 + x + x2)n in ascending powers of x, then prove that : (a)
a0a1 – a1a 2 + a2a3 – .... = 0
(b)
a0a2 – a 1a 3+ a 2a 4 – ....... + a 2n-2 a 2n = a n+1 or a n-1
(c)
E1 = E 2 = E 3 = 3 n-1 ; where E 1 = a 0+ a 3+a 6+ ....... ; E 2 = a 1 + a 4+ a 7 +....... & E 3 = a 2+ a 5+a 8+.......
5.
Prove that : 12 . C0 + 22 C1 + 32. C2 + 42. C3. +........+(n +1)2 . Cn= 2n-2 ( n +1) (n + 4).
6.
2 3 4 n 2 r 3 n 2 2n 5 If (1+ x)n = C r .x then prove that ; 2 .C 0 2 .C 1 2 .C 2 ....... 2 .C n r 0 1.2 2.3 3.4 (n 1)(n 2) (n 1)(n 2)
7.
r ni n r 1 n Prove that : k 1 k 1 i 0 k
8.
p q p q Prove that : , p, q N; p, q are constants. p n j 0 j n j
9.
n n 1 n 2n 1 Prove that : n r r n 1 r 1
10.
Prove that :
C 0 C1 C 2 C 3 C 1 n. 2 n 1 ....... n 2 3 4 5 n 2 (n 1)(n 2)
11.
Prove that :
1 n 2 3 4 ( 1) n 1 n n 1 C 1 n C 2 n C 3 n C 4 ..... . Cn 2 3 4 5 n 1 n 1
12.
Prove that : (2nC 1)2 +2. (2nC 2)2 +3.( 2nC3)2 +..... +2n.(2nC 2n) 2 =
NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#11\Eng\01.BINOMIAL.p65
n
E
BRAIN STOR MIN G SUBJ ECTIVE E X ER CISE
3.
(a)
n = 12
(b)
ANSWER
5 20 x 8 21
23
(4n 1)! [(2n 1)!]2
KEY
EXERCISE-4(B)
JEE-Mathematics
Page No: 24
EXERCISE - 05 [A]
The sum of the coefficients in the expansion of (x + y)n is 4096. The greatest coefficient in the expansion is[AIEEE 2002] (1) 1024
2.
(2) 924
n n 1 2
n n 2 2
n n 1 256
n 1 n 2
(4)
2
3 85
(2) 33
2
is-
(3) 34
5 3
(2)
10 3
(3)
[AIEEE
3 10
3 5
(4)
(2) (–1) n (1–n)
[AIEEE
(3) (–1) n–1 (n –1) 2
10.
If the coefficients of r th, (r + 1) th and (r + 2) th terms in the binomial expansion (1 + y) m are in A.P., then m and r satisfy the equation[AIEEE 2005] (1) m 2 – m (4r – 1) + 4r 2 + 2 = 0
(2)
(3) m 2 – m (4r + 1) + 4r 2 + 2 = 0
(4) m 2 – m(4r – 1) + 4r 2 – 2 = 0
1 If the coefficient of x7 in ax 2 bx
m 2 – m (4r + 1) + 4r 2 – 2 = 0
11
1 equals the coefficient of x–7 in ax bx 2
, then a and b satisfy [AIEEE
2005]
a 1 (3) a + b = 1 (4) a – b = 1 b For natural numbers m, n if (1 – y) m (1 + y) n = 1 + a 1 y + a 2 y 2 +......, and a 1 = a 2 = 10, then (m, n) [AIEEE 2006] is-
(1) ab = 1
(2)
(1) (45, 35)
(2) (35, 45)
The sum of the series 20 C (1)
11.
2004]
(4) (–1) n–1 n
the relation-
9.
2003]
(4) 35
The coefficient of x n in expansion of (1 + x)(1 – x) n is-
11
8.
(3)
[AIEEE-2002]
The coefficient of the middle term in the binomial expansion in powers of x of (1 + x) 4 and of (1 – x) 6 is the same if equals[AIEEE 2004]
(1) (n – 1) 7.
(2)
(4) n = 2r – 1
C1 2C 2 3C 3 nC n ... C0 C1 C2 C n 1
The number of integral terms in the expansion of
(1) 6.
(3) n = 2r + 1
n
(1) 32 5.
(2) n = 3r
If 1 x C 0 C 1 x C 2 x 2 ... ...C n x n , then (1)
4.
(4) 724
If for positive integers r > 1, n > 2 the coefficients of the (3r) th and (r+2) th powers of x in the expansion [AIEEE 2002] of (1+x) 2n are equal, then(1) n = 2r
3.
(3) 824
1 20 C 10 2
0
20 C
1
+ 20 C
(3) (20, 45) 2
20 C
3
+ ..... ...... + 20 C (3) – 20 C 10
(2) 0
(4) (35, 20) 10 is -
[AIEEE 2007]
(4) 20 C 10
In the binomial expansion of (a b) n , n 5, the sum of 5 th and 6 th terms is zero, then
a equals b
[AIEEE
(1)
6 n 5
(2)
n 5 6
(3)
24
n4 5
(4)
5 n4
2007]
NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#11\Eng\01.BINOMIAL.p65
1.
JEE-[MAIN] : PREVIOUS YEAR QUESTIONS
E
JEE-Mathematics
Page No: 25 n
12.
r 1 C n 2 2 n
Statement –1 :
n 1
r
r 0 n
Statement–2 : r 1 n C r x r = (1 + x)n+nx (1+x) n–1
[AIEEE 2008]
r 0
(1) Statement –1 is false, Statement –2 is true (2) Statement–1 is true, Statement–2 is true; Statement–2 is a correct explanation for Statement–1 (3) Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for Statement–1 (4) Statement–1 is true, Statement–2 is false 13.
The remainder left out when 8 2n – (62) 2n+1 is divided by 9 is :(1) 7
(2) 8 10
14.
[AIEEE 2009]
(3) 0
10
10
(4) 2
10
Let S 1 j( j 1)10 C j , S 2 j10 C j and S 3 j2 C j . j 1
j 1
[AIEEE-2010]
j 1
Statement–1 : S 3 = 55 × 2 9 . Statement–2 : S 1 = 90 × 2 8 and S 2 = 10 × 2 8 . (1) Statement–1 is true, Statement–2 is true; Statement–2 is a correct explanation for Statement–1. (2) Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for statement–1. (3) Statement–1 is true, Statement–2 is false. (4) Statement–1 is false, Statement–2 is true. 15.
The coefficient of x 7 in the expansion of (1 – x – x 2 + x 3)6 is :(1) –144
16.
(2) 132
[AIEEE 2011]
(3) 144 2n
If n is a positive integer, then 3 1
2n
3 1
(4) – 132
is :
[AIEEE 2012]
(1) a rational number other than positive integers
(2) an irrational number
(3) an odd positive integer
(4) an even positive integer 10
NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#11\Eng\01.BINOMIAL.p65
17.
E
x 1 x 1 The term independent of x in expansion of 2/3 1/3 x x 1 x x1 / 2 (1) 4
(2) 120
(3) 210
ANSWER
P RE VIOU S Y EARS QU E STION S
Q ue.
1
is :
[JEE
(Main)-2013]
(4) 310
KEY
E XE R CISE -5
[A]
2
3
4
5
6
7
8
9
10
11
12
13
14
15
3
2
3
2
2
1
2
1
3
2
4
3
1
Ans
2
3
Q ue.
16
17
Ans
2
3
25
JEE-Mathematics
Page No: 26
EXERCISE - 05 [B] 1.
JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS
n n n ( a ) For 2 rn , + 2 + = r r 1 r 2 n 1 (A) r 1
[JEE 2000, (Screening ), 1+1M]
n 1 (B) 2 r 1
n 2 (C) 2 r
n 2 (D) r
a equals b
( b ) In the binomial expansion of (a - b)n, n 5, the sum of the 5th and 6th terms is zero, Then
n5 6
(A) 2.
(B)
n4 5
5 n4
(C)
n For any positive integers m , n (with n m) , let = m
(D)
6 n5
n
Cm . Prove that :
n n 1 n 2 n 1 m m + m + m + ........ + = m 1 m
Hence or otherwise prove that, n n 1 m + 2 m
n 2 + 3 m
10 20
H qK
i 0
(A) 5 4.
m n 2 + ........ + (n m + 1) = . m m 2
F pI The sum (where G J 0 ) if p < q is maximum when m is i m i m
3.
[JEE 2000 (Mains), 6M]
(B) 10
( a ) Coefficient of t (A)
12
24
(C) 15 2 12
in the expansion of (1 + t )
C6 + 2
(B)
12
[JEE 02(Screening ), 3M]
12
(D) 20 24
(1 + t ) (1 + t ) is -
C6 + 1
12
(C)
[JEE 03, Screening , 3M out of 60]
C6
(D) none
( b ) If n and k are positive integers, show that
[JEE 03, Mains 2M out of 60]
n n n n 1 n n 2 n n k n 2 k 2 k 1 2 k 2 .....( 1) k 0 k 1 k 1 2 k 2 k 0 k 5.
If n, r N and n–1Cr = (k2 – 3) (nCr+1), then k lies in the interval [JEE 04, Screeni ng, 3M out of 84]
6.
3
The value of
(C) 3,
(B) (2, )
(D)
3, 2
FG30IJ FG 30IJ FG 30IJ FG 30 IJ FG 30IJ FG 30 IJ ..........FG 30 IJ FG30 IJ , is where FG nIJ C Hr K H 0 K H10 K H 1 K H 11K H 2 K H 12K H 20K H 30K n
r
[JEE 05, Screeni ng, 3M out of 84] 30 C
7.
31 C (A) (B) (C) (D) 30C11 10 20 11 or 10 r 10 For r = 0, 1,....,10, let Ar, Br and Cr denote, respectively, the coefficient of x in the expansions of (1+x) ,
(1 + x)
20
60 C
30
31 C
10
and (1 + x) . Then
A (B B C A ) is equal to r
10
r
10
[JEE 10, 5M, –2M]
r
r 1
8.
2
(B) A 10 B 10 C 10 A 10
(A) B 10 – C 10
(C) 0
The coefficients of three consecutive terms of (1 + x)
(D) C 10 – B 10
n+5
are in the ratio 5 : 10 : 14. Then n = [JEE-Advanced 2013, 4, (–1)]
P RE VIOU S Y EARS QU E STION S
1. (a) D ; (b) B
3. C
4. (a) A
ANSWER
KEY
5. D
6. A
26
E XE R CISE -5
7.
D
8.
6
[B]
NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#11\Eng\01.BINOMIAL.p65
(A) 3,
E