hw5-sol - k IS: Suppose we have a pile of k + 1 stones....

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Rob Bayer Math 55 Homework 5 SOLUTIONS July 14, 2009 § 4 . 1#10 (a) Find a formula for 1 1 · 2 + 1 2 · 3 + ··· + 1 n ( n + 1) The first few terms are 1 2 , 2 3 , 3 4 ,... so we guess n n +1 (b) We’ll go by induction: BC: for n = 1, both sides evaluate to 1 2 IH: Suppose 1 1 · 2 + 2 · 3 + ... + 1 k ( k +1) = k k +1 IS: 1 1 · 2 + 1 2 · 3 + ... + 1 k ( k + 1) + 1 ( k + 1)( k + 2) IH = k k + 1 + 1 ( k + 1)( k + 2) = k ( k + 2) + 1 ( k + 1)( k + 2) = ( k + 1) 2 ( k + 1)( k + 2) = k + 1 k + 2 as was to be shown. § 4 . 2#14 Show that if you start with n stones in a pile an incur a cost of k 1 k 2 whenever you break a pile into two piles of sizes k 1 ,k 2 , then the total cost to break the pile of n down to single stones is n ( n - 1) 2 . We’ll go by strong induction: BC: 1 stone is already broken apart, so the cost is 0, which is the same as 1(1 - 1) 2 IH: Suppose the formula works for piles of size
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Unformatted text preview: k IS: Suppose we have a pile of k + 1 stones. Then if we break it into piles of sizes k 1 , k 2 we incur a cost of k 1 k 2 . By the IH, it costs k 1 ( k 1-1) 2 and k 2 ( k 2-1) 2 to fully break apart each of these. Thus, the total cost is: k 1 k 2 + k 1 ( k 1-1) + k 2 ( k 2-1) 2 = 2 k 1 k 2 + k 2 1-k 1 + k 2 2-k 2 2 = k 2 1 + 2 k 1 k 2 + k 2 2-( k 1 + k 2 ) 2 = ( k 1 + k 2 ) 2 + ( k 1 + k 2 ) 2 = ( k + 1) 2-( k + 1) 2 = ( k + 1)( k + 1-1) 2 as was to be shown. Thus, by strong induction, the result holds for all n ....
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This note was uploaded on 02/06/2012 for the course MATHEMATIC 55 taught by Professor Robbayer during the Summer '09 term at University of California, Berkeley.

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