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Unformatted text preview: n odd the proof is straightforward. There are an even number of terms in the sum (0 , 1 , ,n ), and ( n k ) and ( n nk ) , which are equal, have opposite signs. Thus, all pairs cancel and the sum is zero. If n is even, use the following identity, which is the basis of Pascals triangle: For k > 0, ( n k ) = ( n1 k ) + ( n1 k1 ) . Then, for n even n X k =0 (1) k n k = n + n1 X k =1 (1) k n k + n n = n + n n + n1 X k =1 (1) k n1 k + n1 k1 = n + n n  n1  n1 n1 = 0 . b. Use the fact that for k > 0, k ( n k ) = n ( n1 k1 ) to write n X k =1 k n k = n n X k =1 n1 k1 = n n1 X j =0 n1 j = n 2 n1 ....
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.
 Spring '12
 Dr.Hackney
 Statistics

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