CMPUT272 Lecture Notes - Lecture 10: Natural Number, Chessboard

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Theroem: for all n +, 1^2 + 2^2 + + n^2 = (1/6)n(n+1)(2n+1) n. 1^2 + 2^2 + + n^2 = (i^2) i=1. 1 for n = 1: i^2 = 1^2 = 1 = (1/6)n(n+1)(2n+1) = (1/6)(1)(2)(3) = 1 i=1. Thus, the thereom holds for n = 1. 2 for n = 2: i^2 = 1^2 + 2^2 = 5 = (1/6)n(n+1)(2n+1) = (1/6)(2)(3)(5) = 5 i=1. Thus, the tereom holds for n = 2. Suppose we have verified up to n = k k for n = k: i^2 = = (1/6)(k)(k+1)(2k+1) i=1. How does the sum change as n increases from k, to k + 1? k. I^2 = 1^2 + 2^2 + + k^2 i=1 k+1. I^2 = 1^2 + 2^2 + + k^2 + (k+1)^2 i=1. Thus, by substituting the equivilance back to the summation (use the result from previous case to prove next case) k+1.

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