Example 1 - Chapter 4 Class 11 Mathematical Induction (Deleted)

Transcript

Example 1 For all n ≥ 1, prove that 12 + 22 + 32 + 42 +…+ n2 = (n(n+1)(2n+1))/6 Let P (n) : 12 + 22 + 32 + 42 + …..+ n2 = (n(n+1)(2n+1))/6 For n = 1, L.H.S = 12 = 1 R.H.S = (1(1+1)(2 × 1+ 1))/6 = (1 × 2 × 3)/6 = 1 Hence, L.H.S. = R.H.S ∴ P(n) is true for n = 1 Assume that P(k) is true 1 + 22 + 32 +… …+ k2 = (k (k + 1)(2k + 1))/6 We will prove that P(k + 1) is true. 1 + 22 + 32 +… …+ (k + 1)2 = ((k + 1)((k + 1)+ 1)(2 × (k + 1) +1))/6 1 + 22 + 32 +… …+ (k + 1)2 = ((k + 1)(k + 2)(2k + 2 +1))/6 1 + 22 + 32 +… …+ k2 + (k + 1)2 = ((k + 1)(k + 2)(2k + 3))/6 We have to prove P(k+1) from P(k) i.e. (2) from (1) From (1) 1 + 22 + 32 +… …+ k2 = (k (k + 1)(2k + 1))/6 Adding (k+1)2 both sides, 12 +22 +32 +42 +…+k2 + (k + 1)2 = (k (k + 1)(2k + 1))/6 + (k + 1)2 = (k(k + 1)(2k + 1) + 6(k + 1)2)/6 = ((k + 1)(k(2k + 1) + 6(k + 1)))/6 = ((k + 1)(2k2 + k + 6k + 6))/6 = ((k + 1)(2k2 + 7k + 6))/6 = ((k + 1)(2k2 + 4k + 3k + 6))/6 = ((k + 1)(2k(k + 2) + 3(k + 2)))/6 = ((k + 1)(2k + 3)(k + 2))/6 Thus, 12 +22 +32 +42 +…+k2 + (k + 1)2 = ((k + 1)(2k + 3)(k + 2))/6 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