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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.