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Unformatted text preview: CSE 21: Solutions  Problem Set 2 1. (a) x = 1, y = 1 (b) x = 1, y = 1. n โ k =0 ( 1) k ยท 1 n k n k = ( 1 + 1) n โ โ k even n k X k odd n k = โ โ k even n k = X k odd n k (c) x = 2, y = 1. (d) x = i , y = 1, where i is the imaginary number โ 1. Let n = 4 m . 4 m X k =0 4 m k i k ยท 1 4 m k = ( i + 1) 4 m 4 m X k =0 4 m k i k = [( i + 1) 2 ] 2 m X k even 4 m k i k + X k odd 4 m k i k = ( i 2 + 2 i + 1) 2 m 2 m X k =0 4 m 2 k i 2 k + 2 m 1 X k =0 4 m 2 k + 1 i 2 k +1 = (2 i ) 2 m 2 m X k =0 4 m 2 k ( 1) k + 2 m 1 X k =0 4 m 2 k + 1 ( 1) k i = ( 4) m Take the real parts on both sides, we have the desired identity. Note : If we take the imaginary parts, we have 2 m 1 X k =0 ( 1) k 4 m 2 k + 1 = 0 , which is also obvious by the symmetry of the binomial coefficients for n = 4 m . 2. (a) If no restriction, this is equivalent to the Balls and Bins problem. The answer is 25 + 9 1 9 1 = 33 8 (b) We can first give one jellybean to each boy and 2 jellybeans to each girl (only one way to do this since jellybeans are undistinguishable), then distribute the remaining jellybeans with no...
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This note was uploaded on 04/23/2008 for the course CSE 21 taught by Professor Graham during the Spring '07 term at UCSD.
 Spring '07
 Graham

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