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Unformatted text preview: (3) = 3 2 = 9 s (4) = 3 2 + 2 · 3 3 = 9 + 54 = 63 s (5) = 3 2 + 2 · 3 3 + 3 · 3 4 = 9 + 54 + 243 = 306 (e) (2 points) Find a closed form for ∑ n k =1 k 3 . Hint: look it up in a table of closed forms for summations. Solution: n X k =1 k 3 = ( n ( n + 1) / 2) 2 1 (f) (4 points) Using your result from part (e), ﬁnd closed forms for ∑ n k =1 k 3 ( n1) and ∑ n k =2 ( n + k 3 ) . Solution: n X k =1 k 3 ( n1) = ( n1) · n X k =1 ( k 3 ) = ( n1) · ( n ( n + 1) / 2) 2 n X k =2 ( n + k 3 ) = n X k =1 ( n + k 3 )( n + 1 3 ) = n X k =1 n + n X k =1 ( k 3 )( n + 1) = n 2 + ( n ( n + 1) / 2) 2( n + 1) 2...
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This note was uploaded on 09/19/2011 for the course CS cs173 taught by Professor Fleck during the Fall '07 term at University of Illinois, Urbana Champaign.
 Fall '07
 fleck

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