Adv Alegbra HW Solutions 15

Adv Alegbra HW Solutions 15 - 15 1.32 (i) Show, for every n...

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Unformatted text preview: 15 1.32 (i) Show, for every n ≥ 1, that the “alternating sum” of the binomial coefficients is zero: n n n n = 0. − + − · · · + (−1)n n 0 1 2 Solution. If f (x ) = (1 + x )n , then f (−1) = (1 − 1)n = 0; but the expansion is the alternating sum of the binomial coefficients. (ii) Use part (i) to prove, for a given n , that the sum of all the binomial n n coefficients r with r even is equal to the sum of all those r with r odd. Solution. By part (i), n n n n − + − ··· ± = 0. 0 1 2 n Since the sign of n is (−1)r , the terms n with r even are posir r tive while those with r odd are negative. Just put those coefficients with negative coefficient on the other side of the equation. 1.33 Prove that if n ≥ 2, then n (−1)r −1r r =1 n = 0. r Solution. Again, consider f (x ) = (1+ x )n . There are two ways to describe its derivative f (x ). On the one hand, f (x ) = n (1 + x )n −1 . On the other hand, we can do term-by-term differentiation: n f (x ) = r r =1 n r −1 x. r Evaluating at x = −1 gives f (−1) = n (1 − 1)n −1 = 0, since n − 1 ≥ 1. On the other hand, we can use the expansion to see n f (−1) = (−1)r −1r r =1 1.34 If 1 ≤ r ≤ n , prove that n n−1 n = . r r r −1 Solution. Absent. n . r ...
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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