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Adv Alegbra HW Solutions 15

Adv Alegbra HW Solutions 15 - 15 1.32(i Show for every n 1...

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15 1.32 (i) Show, for every n 1, that the alternating sum of the binomial coef fi cients is zero: n 0 n 1 + n 2 − · · · + ( 1 ) n n n = 0 . Solution. If f ( x ) = ( 1 + x ) n , then f ( 1 ) = ( 1 1 ) n = 0; but the expansion is the alternating sum of the binomial coef fi cients. (ii) Use part (i) to prove, for a given n , that the sum of all the binomial coef fi cients ( n r ) with r even is equal to the sum of all those ( n r ) with r odd. Solution. By part (i), n 0 n 1 + n 2 − · · · ± n n = 0 . Since the sign of ( n r ) is ( 1 ) r , the terms ( n r ) with r even are posi- tive while those with r odd are negative. Just put those coef fi cients with negative coef fi cient on the other side of the equation. 1.33 Prove that if n 2, then n r = 1 ( 1 ) r 1 r n r =
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