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Unformatted text preview: MAD 6206 Combinatorics I, Fall 2011 CRN: 87401 Assignment 6 – suggested solution Question 1 . (Two wellknown series in hypergeometric notation) (a) Consider the sum ∑ k ( n + k k ) z k . We know how to find a closed form from other methods. In particular, the sum follows from the generalized binomial theorem since ( n + k k ) = ( n 1 k ) ( − 1) k = (1 − z ) n 1 . Put the sum into hypergeometric notation, and show that 1 F bracketleftbigg a − ; z bracketrightbigg = (1 − z ) a . (b) Do the same for the sum ∑ k ≥ 1 k ! z k . Solution . (a) Note first that the first term is t = ( n ) = 1. Also, t k +1 t k = n + k + 1 k + 1 z. Hence, the sum is 1 F bracketleftbigg n + 1 , − ; z bracketrightbigg . As the sum equals (1 − z ) n 1 , we therefore have 1 F bracketleftbigg a − ; z bracketrightbigg = (1 − z ) a . (b) For ∑ k ≥ 1 k ! z k = e z , we have t k +1 t k = 1 k + 1 . Thus, in the hypergeometric notation, the sum is F bracketleftbigg − − ; z bracketrightbigg = e z . Question 2 . (Reduction to hypergeometric notation) Consider the sum ∑ k ( 2 n +2 2 k +1 )( x + k 2 n +1 ) . (a) Reduce the sum into hypergeometric notation. (b) Assume that n ≥ 0 is an integer. Look up a hypergeometric identity for the sum, and show that summationdisplay k parenleftbigg 2 n + 2 2 k + 1 parenrightbiggparenleftbigg x + k 2 n + 1 parenrightbigg = 2( x − n ) 2 n + 1 parenleftbigg 2 x 2 n parenrightbigg . Solution . Observe that t = parenleftbigg 2 n + 2 1 parenrightbiggparenleftbigg x 2 n + 1 parenrightbigg ....
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This note was uploaded on 01/02/2012 for the course MAD 6206 taught by Professor Suen during the Spring '11 term at University of South Florida  Tampa.
 Spring '11
 Suen

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