Ex 4.1, 15 - Prove 12 + 32 + 52 ..+ (2n-1)2 - Chapter 4 Induction - Ex 4.1

  1. Chapter 4 Class 11 Mathematical Induction
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Ex 4.1,15 Prove the following by using the principle of mathematical induction for all n ∈ N: 12 + 32 + 52 + …..+ (2n – 1)2 = (n(2n − 1)(2n + 1))/3 Let P (n) : 12 + 32 + 52 + …..+(2n – 1)2 = (n(2n − 1)(2n + 1))/3 For n = 1, L.H.S = 12 = 1 R.H.S = (1(2 × 1 − 1)(2 × 1+ 1))/3 = (1(2 − 1) (2 + 1))/3 = (1 × 1 × 3)/3 = 1 Hence L.H.S. = R.H.S ∴P(n) is true for n = 1 Assume that P(k) is true 12 + 32 + 52 + …..+ (2k – 1)2 = (k(2k − 1)(2k + 1))/3 We will prove that P(k + 1) is true. 12 + 32 + 52 + …+ (2(k + 1) – 1)2 = ( (k + 1)(2(k + 1)− 1)(2(k + 1)+ 1))/3 12 + 32 + 52 +…+ (2k – 1)2 + (2k + 2 – 1)2 = ((k + 1)(2k + 2 − 1)(2k + 2 + 1))/3 12 + 32 + 52 +… + (2k – 1)2 + (2k + 1)2 = ((k + 1)(2k + 1)(2k + 3))/3 From (1) 12 + 32 + 52 + …..+ (2k – 1)2 = (k(2k − 1)(2k + 1))/3 Adding (2k + 1)2 both sides 12 + 32 + 52 + …..+ (2k – 1)2 + (2k + 1)2 = (k(2k − 1)(2k + 1))/3 + (2k + 1)2 = (k(2k − 1)(2k + 1) + 3(2𝑘 + 1)^2)/3 = (2k + 1)((k(2k − 1)+ 3(2𝑘 + 1))/3) = (2k + 1)((k(2k) −k(1)+ 3(2𝑘) +3(1))/3) = (2k + 1)((2k^2 − 𝑘 + 6𝑘 + 3)/3) = (2k + 1)((2k^2 + 5𝑘 + 3)/3) = (2k + 1)((2k^2 + 2𝑘 + 3𝑘 + 3)/3) = (2k + 1)((2k(k + 1) + 3(𝑘 +1))/3) = (2k + 1)(((2k + 3)(𝑘 +1))/3) = ((k + 1)(2k + 1)(2k + 3))/3 Thus, 12 + 32 + 52 +… + (2k – 1)2 + (2k + 1)2 = ((k + 1)(2k + 1)(2k + 3))/3 which is the same as P(k + 1) ∴ P(k+1) is true when P(k) is true ∴ By the principle of mathematical induction, P(n) is true for n, where n is a natural number

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