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m25-hw2-solutions

# m25-hw2-solutions - Math 25 Advanced Calculus Fall 2010...

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Math 25: Advanced Calculus Fall 2010 Homework # 2 – Solutions Due Date: Friday, October 8 A.8.1: Prove by induction that for every n = 1 , 2 , 3 , . . . , 1 2 + 2 2 + · · · + n 2 = n ( n + 1)(2 n + 1) 6 . Proof: First we establish the base case n = 1, where it is clear that 1 2 = 1 = 1 · 2 · 3 6 . For the inductive step, we assume that the desired equation holds for an integer n - 1: 1 2 + 2 2 + · · · + ( n - 1) 2 = ( n - 1)( n )(2 n - 1) 6 , and we show that the equation also holds for the integer n . To do this, we compute 1 2 + 2 2 + · · · + n 2 = 1 2 + 2 2 + · · · + +( n - 1) 2 + n 2 = ( n - 1)( n )(2 n - 1) 6 + n 2 by our induction hypothesis = 1 6 ( ( n - 1)( n )(2 n - 1) + 6 n 2 ) = 1 6 · n · ( 2 n 2 - 3 n + 1 + 6 n ) = 1 6 · n · ( 2 n 2 + 3 n + 1 ) = 1 6 · n · ((2 n + 1)( n + 1)) = n ( n + 1)(2 n + 1) 6 , as desired. Therefore, by the principle of mathematical induction, we have established the desired claim. 1

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A.8.4: Prove by induction that (1 + x ) n 1 + nx, for any integer x > 0 and any n = 1 , 2 , 3 , . . . . Proof: First we establish the base case n = 1. This holds since (1 + x ) 1 = 1 + 1 · x for any x . For the inductive step, we assume that the desired equation holds for an integer n - 1, i.e. that (1 + X ) n - 1 1 + ( n - 1) x for any x > 0, and we show that the equation also holds for the integer n . To do this, we compute: (1 + x ) n = (1 + x )(1 + x ) n - 1 (1 + x )(1 + ( n - 1) x ) by our induction hypothesis = 1 + nx + ( n - 1) x 2 1 + nx since ( n - 1) x 2 0 . Therefore, by the principle of mathematical induction, we have established the desired claim. 2
A.9.1: Express in words what these statements mean, and determine whether they are true or not. (a) x R , x 0 For all real numbers x , x is greater than or equal to zero. (FALSE) (b) x R , x 0.

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m25-hw2-solutions - Math 25 Advanced Calculus Fall 2010...

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