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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 k1) 30 + ( k + 1) 4 = 6 k 5 +15 k 4 +10 k 3k +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).
 Spring '08
 BILLERA
 Math, Algebra

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