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Unformatted text preview: Spring 2011 Solutions to Homework 2 MAT 25 1. Prove by induction that for every n N 1 2 + 2 2 + + n 2 = n ( n + 1)(2 n + 1) 6 . Proof. When n = 1, the statement becomes 1 = 1 2 = 1(1 + 1)(2 + 1) 6 = 6 6 = 1 , which is clearly true. Now, assume that 1 2 + 2 2 + + n 2 = n ( n + 1)(2 n + 1) 6 . We want to show that 1 2 + 2 2 + + n 2 + ( n + 1) 2 = ( n + 1)( n + 2)(2 n + 3) 6 . Now, 1 2 + 2 2 + + n 2 + ( n + 1) 2 = 1 2 + 2 2 + + n 2 + ( n + 1) 2 = n ( n + 1)(2 n + 1) 6 + ( n + 1) 2 = n ( n + 1)(2 n + 1) 6 + 6 n 2 + 12 n + 6 6 = n ( n + 1)(2 n + 1) + 6 n 2 + 12 n + 6 6 = 2 n 3 + 3 n 2 + n + 6 n 2 + 12 n + 6 6 = 2 n 3 + 7 n 2 + 6 n + 2 n 2 + 7 n + 6 6 = n (2 n 2 + 7 n + 6) + (2 n 2 + 7 n + 6) 6 = ( n + 1)(2 n 2 + 7 n + 6) 6 = ( n + 1)( n + 2)(2 n + 3) 6 , as desired. 2. Let x R and x > 0. Prove by induction that for all n N (1 + x ) n 1 + nx. Proof. When n = 1, the statement is simply 1 + x 1 + x , which is clearly true. So, assume (1 + x ) n 1 + nx. Then, we want to show (1 + x ) n +1 1 + ( n + 1) x. Now, (1 + x ) n +1 = (1 + x )(1 + x ) n (1 + x )(1 + nx ) = 1 + nx + x + nx 2 = 1 + ( n + 1) x + nx 2 1 + ( n + 1) x, where in the first inequality in this chain of inequalities we used the inductive hypothesis together with the fact that...
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This note was uploaded on 10/16/2011 for the course MATH 34 taught by Professor Wiley during the Winter '11 term at UC Merced.
 Winter '11
 Wiley
 Differential Equations, Equations

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