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Examples

Example 1
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Example 8 Deleted for CBSE Board 2023 Exams

Last updated at Jan. 5, 2022 by Teachoo

Example 1 For all n β₯ 1, prove that 12 + 22 + 32 + 42 +β¦+ n2 = (n(n+1)(2n+1))/6 Let P(n) : 12 + 22 + 32 + 42 + β¦..+ n2 = (π(π + 1)(2π + 1))/6 Proving for n = 1 For n = 1, L.H.S = 12 = 1 R.H.S = (1(1+1)(2 Γ 1+ 1))/6 = (1 Γ 2 Γ 3)/6 = 1 Since, L.H.S. = R.H.S β΄ P(n) is true for n = 1 Proving P(k + 1) is true if P(k) is true Assume that P(k) is true, P(k): 1 + 22 + 32 +β¦ β¦+ k2 = (π (π + 1)(2π + 1))/6 We will prove that P(k + 1) is true. P(k + 1): 1 + 22 + 32 +β¦ β¦+ (k + 1)2 = ((π + 1)((π + 1)+ 1)(2 Γ (π + 1) +1))/6 P(k + 1): 1 + 22 + 32 +β¦ β¦+ (k + 1)2 = ((π + 1)(π + 2)(2π + 2 +1))/6 P(k + 1): 1 + 22 + 32 +β¦ β¦+ k2 + (k + 1)2 = ((π + π)(π + π)(ππ + π))/π We have to prove P(k + 1) is true Solving LHS 1 + 22 + 32 +β¦ β¦+ k2 + (k + 1)2 From (1): 1 + 22 + 32 +β¦ β¦+ k2 = (π (π + 1)(2π + 1))/6 = (π (π + π)(ππ + π))/π + (k + 1)2 = (π(π + 1)(2π + 1) + 6(π + 1)2)/6 = ((π + 1)(π(2π + 1) + 6(π + 1)))/6 = ((π + 1)(2π2 + π + 6π + 6))/6 = ((π + π)(πππ + ππ + π))/π = ((π + 1)(2π2 + 4π + 3π + 6))/6 = ((π + 1)(2π(π + 2) + 3(π + 2)))/6 = ((π + π)(ππ + π)(π + π))/π = RHS β΄ P(k + 1) is true when P(k) is true Thus, By the principle of mathematical induction, P(n) is true for n, where n is a natural number