quiz5aANS - Base Step: n = 1. Then n 3-n = 0, and 6 | 0....

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 240, Quiz 5a Name: Circle One: T 12:05 T 2:25 R 12:05 R 2:25 Instructions: Answer all questions fully, showing work where necessary. 1) Compute each of these sums: a) 8 j =0 (2 · 3 j + 3 · 2 j ) People made this problem a lot harder than it needs to be. We can separate this out into 8 j =0 2 · 3 j + 8 j =0 3 · 2 j . These are both geometric sums. So 8 j =0 2 · 3 j = 2 · 3 9 - 2 3 - 1 = 19682. Also, 8 j =0 3 · 2 j = 3 · 2 9 - 3 2 - 1 = 1533. Add these together to get 21215. b) 3 i =1 3 j =1 ( i - j ) Just write out the terms, and you’ll see they add to 0. 2) Use mathematical induction to show that 6 divides n 3 - n whenever n is a nonnegative integer. Proof by Induction:
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Base Step: n = 1. Then n 3-n = 0, and 6 | 0. Induction step: Assume 6 | n 3-n . We now need to show 6 | ( n +1) 3-( n +1), or that 6 | n 3 + 3 n 2 + 3 n + 1-n-1. Simplifying, we get ( n + 1) 3-( n + 1) = n 3-n + 3 n 2 + 3 n . By the inductive hypothesis, 6 | n 3-n . Thus, we need to show 6 | 3 n 2 + 3 n , or 6 | 3 n ( n + 1). Note that 3 | 3 n so all we need to show is that n ( n + 1) is divisible by 2. But one of n and n + 1 must be even, so 3 n ( n + 1) must be divisible by 6. 1...
View Full Document

This note was uploaded on 08/11/2008 for the course MATH 240 taught by Professor Miller during the Fall '08 term at Wisconsin.

Ask a homework question - tutors are online