Check sibling questions

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Example 1 For all n β‰₯ 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 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 ∴ 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 +… …+ k2 = (π‘˜ (π‘˜ + 1)(2π‘˜ + 1))/6 We will prove that P(k + 1) is true. P(k + 1): 1 + 22 + 32 +… …+ (k + 1)2 = ((π‘˜ + 1)((π‘˜ + 1)+ 1)(2 Γ— (π‘˜ + 1) +1))/6 P(k + 1): 1 + 22 + 32 +… …+ (k + 1)2 = ((π‘˜ + 1)(π‘˜ + 2)(2π‘˜ + 2 +1))/6 P(k + 1): 1 + 22 + 32 +… …+ k2 + (k + 1)2 = ((π’Œ + 𝟏)(π’Œ + 𝟐)(πŸπ’Œ + πŸ‘))/πŸ” We have to prove P(k + 1) is true Solving LHS 1 + 22 + 32 +… …+ k2 + (k + 1)2 From (1): 1 + 22 + 32 +… …+ k2 = (π‘˜ (π‘˜ + 1)(2π‘˜ + 1))/6 = (π’Œ (π’Œ + 𝟏)(πŸπ’Œ + 𝟏))/πŸ” + (k + 1)2 = (π‘˜(π‘˜ + 1)(2π‘˜ + 1) + 6(π‘˜ + 1)2)/6 = ((π‘˜ + 1)(π‘˜(2π‘˜ + 1) + 6(π‘˜ + 1)))/6 = ((π‘˜ + 1)(2π‘˜2 + π‘˜ + 6π‘˜ + 6))/6 = ((π’Œ + 𝟏)(πŸπ’ŒπŸ + πŸ•π’Œ + πŸ”))/πŸ” = ((π‘˜ + 1)(2π‘˜2 + 4π‘˜ + 3π‘˜ + 6))/6 = ((π‘˜ + 1)(2π‘˜(π‘˜ + 2) + 3(π‘˜ + 2)))/6 = ((π’Œ + 𝟏)(πŸπ’Œ + πŸ‘)(π’Œ + 𝟐))/πŸ” = RHS ∴ 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

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.