# HW#1 - SOLUTIONS TO HOMEWORK 1 Problem 1.1 We have to prove...

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SOLUTIONS TO HOMEWORK # 1 Problem 1.1. We have to prove that the statement P n : 1 2 + 2 2 + . . . + n 2 = 1 6 n ( n + 1)(2 n + 1) holds for all n = 1 , 2 , . . . . Step 1 (”base of induction”). P 1 : 1 2 = 1 6 · 1(1 + 1)(2 · 1 + 1) is true. Step 2 . Assume P n holds for some n = k N , that is 1 2 + 2 2 + . . . + k 2 = 1 6 k ( k + 1)(2 k + 1) (1) We will prove that then that it holds for n = k + 1 that is P k +1 : 1 2 + 2 2 + . . . + ( k + 1) 2 = 1 6 ( k + 1)[( k + 1) + 1][2( k + 1) + 1] holds. If (1) holds then 1 2 + 2 2 + . . . + k 2 + ( k + 1) 2 = 1 6 k ( k + 1)(2 k + 1) + ( k + 1) 2 = = k + 1 6 [ k (2 k + 1) + 6( k + 1)] = = 1 6 ( k + 1)[2 k 2 + k + 6 k + 6] = 1 6 ( k + 1)( k + 2)(2 k + 3) = = 1 6 ( k + 1)[( k + 1) + 1][2( k + 1) + 1] . So if P n holds for n = k , then it holds for n = k + 1. By the principle of mathematical induction, P n is true for all n N . Problem 1.6. P n : (11) n - 4 n is divisible by 7. This means (11) n - 4 n = 7 s for some s N . Step 1 (”base of induction”). P 1 : 11 - 4 = 7 · 1, so P 1 holds. Step 2 (”transition”). Assume P n holds for some k N that is (11) k - 4 k = 7 l where l N . (2) We have to prove that it holds for n = k + 1, that is (11) k +1 - 4 k +1 = 7 m, where m N .

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