Principle of Induction

CLASS XI: PROBLEMS IN MATHEMATICAL INDUCTION PROF. SEBASTIAN VATTAMATTAM

1. Mathematical Principle of Induction Let P (n) denote a statement about n ∈ N. If (1) P (1) is true, and (2) P (k) is true implies that P (k + 1) is true, then P (n) is true for all n ∈ N. Problem 1.1. Prove that 1 + 3 + ... + (2n − 1) = n2 Solution Let P (n) := 1 + 3 + ... + (2n − 1) = n2 (1) P (1) := 1 = 12 , which is true.

That is P (1) is true.

(2) Suppose P (k) is true. ⇒ 1 + 3 + ... + (2k − 1) = k 2 1 + 3 + ... + (2k − 1) + (2k + 1) = k 2 + 2k + 1 = (k + 1)2 So, P (k + 1) is true.

By induction P (n), n ∈ N is true.

Problem 1.2. Prove that 12 + 22 + ... + n2 = Date: 25-Sept-2010.

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n(n + 1)(2n + 1) 6

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PROF. SEBASTIAN VATTAMATTAM

Solution Let n(n + 1)(2n + 1) 6 (1) P (1) := 12 = 1(1+1)(2×1+1) , which is true.

6 That is P (1) is true.

(2) Suppose P (k) is true.

P (n) := 12 + 22 + ... + n2 =

k(k + 1)(2k + 1) 6 k(k + 1)(2k + 1) + (k + 1)2 12 + 22 + ... + k 2 + (k + 1)2 = 6 k(k + 1)(2k + 1) + 6(k + 1)2 = 6 (k + 1)(k + 2)[2(k + 1) + 1] = 6 So, P (k + 1) is true.

By induction P (n), n ∈ N is true. ⇒ 12 + 22 + ... + k 2 =

Problem 1.3. Prove that 2n > n, n ∈ N. Proof Let P (n) := 2n > n (1) P (1) := 21 > 1, which is true.

(2) Suppose P (k) is true. ⇒ 2k > k 2k+1 = 2 × 2k > 2k ≥ k + 1 ⇒ 2k+1 > k + 1 So, P (k + 1) is true.

By induction P (n), n ∈ N is true.

Problem 1.4. Prove that n(n + 1)(n + 2) is a multiple of 6.

CLASS XI

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Proof Let P (n) := n(n + 1)(n + 2) is a multiple of 6.

(1) P (1) := 1(1 + 1)(1 + 2) = 6, which is a multiple of 6.

(2) Suppose P (k) is true. ⇒ k(k + 1)(k + 2) is a multiple of 6.

Let k(k + 1)(k + 2) = 6m, m ∈ N.....(1) (k+1)(k+2)(k+3) = k(k+1)(k+2)+3(k+1)(k+2) = 6m+3(k+1)(k+2)....(2) (k + 1)(k + 2) is even. Let (k + 1)(k + 2) = 2p, p ∈ N Substituting in (2), (k + 1)(k + 2)(k + 3) = 6m + 3 × 2p = 6(m + p) ⇒ (k + 1)(k + 1 + 1)(k + 1 + 2) is a multiple of 6.

So, P (k + 1) is true.

By induction P (n), n ∈ N is true.

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