Discrete and Foundational Mathematics I quiz 7 Solutions

# Discrete and Foundational Mathematics I quiz 7 Solutions -...

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MATH 187 – Quiz 7 – Key 1) Prove 1 3 + 2 3 + 3 3 + ··· + n 3 = b n ( n + 1) 2 B 2 (1) for all positive integers n . a) Proof by smallest counterexample. So assume otherwise. That is, we as- sume there is a smallest positive integer x such that (1) is false for n = x . Thus 1 3 + 2 3 + 3 3 + + x 3 n = b x ( x + 1) 2 B 2 . (2) Note that x n = 1 because n = 1 renders (1) to be true. Thus x 2 and thus x 1 < x and x 1 is a positive integer. Thus n = x 1 makes (1) true. So we get 1 3 + 2 3 + 3 3 + + ( x 1) 3 = b ( x 1) x 2 B 2 . (3) Now add x 3 to both sides of (3). Thus 1 3 + 2 3 + 3 3 + + ( x 1) 3 + x 3 = b ( x 1) x 2 B 2 + x 3 = x 2 ± p x 1 2 P 2 + x ² = x 2 ± ( x 1) 4 2 + x ² = x 2 ( x 1) 2 + 4 x 4 = x 2 4 ³( x 2 2 x + 1 ) + 4 x ´ = x 2 4 ( x 2 + 2 x + 1 ) = x 2 4 ( x + 1) 2 = b x ( x + 1) 2 B 2 , which contradicts (2). QED.

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b) Proof by induction on n . 1. Base case. Prove (1) is true for n = 1. Note LHS = 1 3 = 1 and RHS = ( 1 · 2 2 ) 2 = 1. So LHS = RHS. Done with base case. 2. Induction Hypothesis. Assume (1) is true for n = k . So we assume 1 3 + 2 3 + 3 3 + ··· + k 3 = b k ( k + 1) 2 B 2 .
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