# 1-2 - SOLUTIONS FOR HOMEWORK 1 n 3 2 2 1.3 Define...

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SOLUTIONS FOR HOMEWORK 1 1.3. Defne Proposition P n : n k =1 k 3 = n 2 ( n + 1) 2 / 4 We shall show, by induction, that P n For any n N . The base is easy to establish For n = 1. Indeed, P 1 states that 1 3 = 1 2 (1 + 1) 2 / 2, which is true. The induction step consists oF proving that P n implies P n +1 . To this end, assume n k =1 k 3 = n 2 ( n + 1) 2 / 4, and show that n +1 k =1 k 3 = ( n + 1) 2 ( n + 2) 2 / 4. We have: n +1 s k =1 k 3 = n s k =1 k 3 + ( n + 1) 3 = n 2 ( n + 1) 2 4 + ( n + 1) 3 = ( n + 1) 2 4 p n 2 + 4( n + 1) P = ( n + 1) 2 ( n + 2) 2 4 . 1.5. We prove the equality n k =0 2 - k = 2 2 - n by induction. The base oF inductions is taken care oF by veriFying the above equality For n = 1: 1 + 2 - 1 = 2 2 - 1 . This is easily checked. The induction step consists oF showing that, iF n k =0 2 - k = 2 2 - n , then n +1 k =0 2 - k = 2 2 - ( n +1) . Use the induction hypothesis to get n +1 s k =0 2 - k = n s k =0 2 - k + 2 - ( n +1) = 2 2 - n + 2 - ( n +1) = 2 2 · 2 - ( n +1) + 2 - ( n +1) = 2 2 - ( n +1) , and we are done. 1.6. Defne Proposition P n : 7 divides 11 n 4 n . We show by induction that P n is true For every n . The base oF induction is easy to veriFy: when n = 1, 7 divides 11 1 4 1 = 7. The inductive step consists oF showing that, iF 7 divides 11 n 4 n , then 7 also divides 11 n +1 4 n +1 .

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1-2 - SOLUTIONS FOR HOMEWORK 1 n 3 2 2 1.3 Define...

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