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Homework 5 Solution

# Homework 5 Solution - MATH 74 HOMEWORK 5(DUE WEDNESDAY...

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MATH 74 HOMEWORK 5 (DUE WEDNESDAY OCTOBER 3) 1. For n N consider the sequences A n = n j =1 j 2 j and B n = ( n - 1)2 n +1 + 2. We have A 1 = 12 1 = 2 = 0 · 2 2 + 2 = B 1 and for any n N we have A n +1 - A n = n +1 j =1 j 2 j - n j =1 j 2 j = ( n + 1)2 n +1 and B n +1 - B n = ( n 2 n +2 + 2) - (( n - 1)2 n +1 + 2) = n 2 n +2 - n 2 n +1 + 2 n +1 = 2 n 2 n +1 - n 2 n +1 + 2 n +1 = n 2 n +1 + 2 n +1 = ( n + 1)2 n +1 . Thus A n +1 - A n = B n +1 - B n for all n N . We conclude that A n = B n for all n N by the sequence equality theorem. 2. For n N let P ( n ) denote the statement f ( n ) ( x ) = ne x + xe x . P (1) is the statement that f ( x ) = e x + xe x , which is true by the product rule. If P ( k ) is true for some k N , that is, if f ( k ) ( x ) = ke x + xe x , we have f ( k +1 ( x ) = d dx f ( k ) ( x ) by definition = d dx ( ke x + xe x ) by hypothesis = ke x + ( e x + xe x ) by the sum and product rule = ( k + 1) e x + xe x so that P ( k + 1) is also true. By the induction principle we conclude that P ( n ) holds for all n N .

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2 MATH 74 HOMEWORK 5 (DUE WEDNESDAY OCTOBER 3) 3. For n N let P ( n ) denote the statement that n j =1 a j na n .
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Homework 5 Solution - MATH 74 HOMEWORK 5(DUE WEDNESDAY...

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