math131Ahw01

math131Ahw01 - Homework #1 Spring 2010 Homework solutions...

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Homework #1 Math 131A Due: Feb. 4 Spring 2010 Prof. Maruskin Homework solutions by: Liem Tran 1.1 Prove 1 2 + 2 2 + ··· + n 2 = 1 6 n ( n + 1)(2 n + 1) for all n N . Base step : P ( n ) : 1 2 + 2 2 + ··· + n 2 = 1 6 n ( n + 1)(2 n + 1) P (1) : LHS 1 2 = 1 RHS 1 6 (1)(1 + 1)(2(1) + 1)) = 1 6 (1)(2)(3) = 1 Inductive step : Assume P ( k ): 1 2 + 2 2 + ··· + k 2 = 1 6 k ( k + 1)(2 k + 1) Prove P ( k + 1) : 1 2 + 2 2 + ··· + k 2 + ( k + 1) 2 = 1 6 ( k + 1)(( k + 1) + 1)(2( k + 1) + 1) = 1 6 ( k + 1)( k + 2)(2 k + 3) Proof : 1 2 + 2 2 + ··· + k 2 + ( k + 1) 2 = 1 6 k ( k + 1)(2 k + 1) + ( k + 1) 2 = 1 6 ( k 2 + k )(2 k + 1) + ( k 2 + 2 k + 1) = 1 6 (2 k 3 + 3 k 2 + k ) + k 2 = 2 k + 1 = 2 6 k 3 + 3 6 k 2 + 1 6 k + k 2 + 2 k + 1 = 2 6 k 3 + 3 6 k 2 + 1 6 k + 6 6 k 2 + 12 6 k + 6 6 = 2 6 k 3 + 9 6 k 2 + 13 6 k + 6 6 = 1 6 (2 k 3 + 9 k 2 + 13 k + 6) = 1 6 ( k 2 + 3 k + 2)(2 k + 3) = 1 6 ( k + 1)( k + 2)(2 k + 3) 1.2 Prove 3 + 11 + ··· + (8 n - 5) = 4 n 2 - n for all

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This note was uploaded on 09/08/2010 for the course MATH 131A at San Jose State University .

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math131Ahw01 - Homework #1 Spring 2010 Homework solutions...

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