06 - Fall 2009 CS131 Combinatorial Structures Homework 6...

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Fall 2009 CS131 – Combinatorial Structures Homework 6 Homework 6, due Oct 27 You must prove your answer to every question. Do not rely only on the homework for exercise: there are several self-check ex- ercises of the easier kind in the book, try to solve them, too! Problem 1. (10pts) Prove the following identity: 1 - 2 · 2 + 3 · 4 - 4 · 8 +···± n · 2 n - 1 = (1 - ( - 2) n (3 n + 1))/9. (The signs are alternating, this determines ± .) [Hint: use mathematical induction.] Solution. Base case: n = 1, then we have just the first term, 1. The right-hand side is (1 - ( - 2) · 4)/9 = (1 + 8)/9. Induction step: assume that the statement holds for n , we prove it for n + 1. We add ( n + 1)( - 2) n to both sides. Then the left-hand side becomes the sum for n + 1. The right-hand side becomes 1 - ( - 2) n (3 n + 1) 9 + ( n + 1)( - 2) n = 1 - ( - 2) n (3 n + 1) + 9( n + 1)( - 2) n 9 . The numerator is 1 + ( - 2) n (6 n + 8) = 1 - ( - 2) n + 1 (3 n + 4) = 1 - ( - 2) n + 1 (3( n + 1) + 1), which is the numerator of the righ-hand side for
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06 - Fall 2009 CS131 Combinatorial Structures Homework 6...

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