math336_HW1sol - (1 4 + 2 4 + ... + k 4 ) + ( k + 1) 4 = (...

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Homework 1: Math 3360 Applicable Algebra p.13 3. Claim: Show that 1 + 2 + 2 2 + ... + 2 n - 1 = 2 n - 1 for every n > 1. Pf: Base case: Let n = 2. Then we have 1+2 = 3 = 2 2 - 1, so the statement is true for n = 2. Inductive Step: Suppose that this statement holds for n = k . Let n = k + 1. So we have (1 + 2 + ... + 2 k - 1 ) + 2 k = 2 k - 1 + 2 k = 2 · 2 k - 1 = 2 k +1 - 1 Thus by induction our statement is true for all n > 1. 4. Claim: Show that for all n 1, 1 4 + 2 4 + ... + n 4 = n ( n + 1)(2 n + 1)(3 n 2 + 3 n - 1) 30 . Pf: Base Case: Let n = 1. Then we have 1 4 = 1 = (1)(2)(3)(5) 30 = 1(1 + 1)(2 · 1 + 1)(3 · 1 2 + 3 · 1 - 1) 30 so our statement holds for n = 1. Inductive Step: Suppose our statement holds for n = k . Suppose n = k + 1. Then we have
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Unformatted text preview: (1 4 + 2 4 + ... + k 4 ) + ( k + 1) 4 = ( k ( k +1)(2 k +1)(3 k 2 +3 k-1) 30 + ( k + 1) 4 = 6 k 5 +15 k 4 +10 k 3-k +30( k +1) 4 30 = 6 k 5 +45 k 4 +130 k 3 +180 k 2 +119 k +30 30 But we also have ( k +1)( k +1+1)(2( k +1)+1)(3( k +1) 2 +3( k +1)-1) 30 = ( k +1)( k +2)(2 k +3)(3 k 2 +9 k +5) 30 = 6 k 5 +45 k 4 +130 k 3 +180 k 2 +119 k +30 30 Thus we have 1 4 + ... + ( k + 1) 4 = ( k + 1)( k + 1 + 1)(2( k + 1) + 1)(3( k + 1) 2 + 3( k + 1)-1) 30 so by induction our statement is true for all n ≥ 1....
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This note was uploaded on 09/05/2011 for the course MATH 3360 taught by Professor Billera during the Spring '08 term at Cornell University (Engineering School).

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