hw1-solutions - (3 = 3 2 = 9 s(4 = 3 2 2 3 3 = 9 54 = 63...

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CS 173: Discrete Structures, Fall 2011 Homework 1 Solutions This homework contains 1 problem worth 15 points. 1. [15 points] Summations (a) (2 points) Rewrite r k =1 12 log( k +4) as a new summation with an index variable named n which runs from -3 to r - 4 . The terms in your new summation should be equal to the terms in the original summation. Solution: r - 4 X n = - 3 12 log( n + 8) (b) (2 points) Rewrite the sum n i =1 i 2 n substituting k + 1 for n . Solution: k +1 i =1 i 2 k +1 (c) (2 points) Break up the sum t +1 p = - 1 ( p · e - p ) into three parts: its first term, its last term, and a summation containing the remaining terms. Solution: Observe t +1 X p = - 1 ( p · e - p ) = - e + t X p =0 ( p · e - p ) + ( t + 1) · e - t - 1 (d) (3 points) Find the values of the sum s ( n ) = n - 1 i =2 3 i ( i - 1) for values of n from 3 to 5. Solution: s
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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 ( n-1) and ∑ n k =2 ( n + k 3 ) . Solution: n X k =1 k 3 ( n-1) = ( n-1) · n X k =1 ( k 3 ) = ( n-1) · ( 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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hw1-solutions - (3 = 3 2 = 9 s(4 = 3 2 2 3 3 = 9 54 = 63...

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