ma017-2010-hw5

# ma017-2010-hw5 - MA017 2010 HOMEWORK 5 Due 2010-11-02 Tue....

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Unformatted text preview: MA017 2010 HOMEWORK 5 Due 2010-11-02 Tue. (1) One can prove ( n k ) + ( n k +1 ) = ( n +1 k +1 ) by “pure thought” as follows. The ( k + 1)-element subsets of { 1 ,...,n +1 } are of two mutually exclusive types: those that contain n +1, and those that do not; the former are in bijection with the k-element subsets of { 1 ,...,n } , the latter with the ( k + 1)-element subsets of { 1 ,...,n } . Give similar proofs for whichever of the following combinatorial identities strike your fancy. Assume, in all formulae where ( n k ) appears, that k ≤ n . (a) ( n k ) = ( n n- k ) (b) ∑ n k =0 ( n k ) = 2 n (c) ∑ n k =0 k ( n k ) = n 2 n- 1 (d) ∑ n k =0 k 2 ( n k ) = n 2 n- 1 + n ( n- 1)2 n- 2 (e) ∑ n k =0 k 3 ( n k ) = n 2 n- 1 + 3 n ( n- 1)2 n- 2 + n ( n- 1)( n- 2)2 n- 3 (f) ∑ k ( n 2 k ) = ∑ k ( n 2 k +1 ) (g) ( 2 n 2 ) = 2 ( n 2 ) + n 2 (h) ( 2 n +2 n +1 ) = ( 2 n n +1 ) + 2 ( 2 n n ) + ( 2 n n- 1 ) (i) ∑ n k =0 (- 1) k ( n k ) = 0 (j) ( n k ) = n k ( n- 1 k- 1 ) (k) ∑ n i...
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ma017-2010-hw5 - MA017 2010 HOMEWORK 5 Due 2010-11-02 Tue....

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