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Theorem: A positive integer n will be the sum of a sequence of consecutive positive integers if and only if n #2^n.

Theorem: A positive integer n will be the sum of a sequence of consecutive positive integers if and only if n #2^n.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I'm not sure where to go from here. We did a problem on number from 1-35 and finding all the ways to find all the possible positive consecutive integer numbers for each and then we have to write a proof on this. And I know that if it's and if and only if statement it needs to be proved both ways but I'm Not sure how to do it. I started it this way

Theorem: A positive integer n will be the sum of a sequence of consecutive positive integers if and only if n #2^n. Proof: Let s be the smallest of c consecutive integers € Z + such that c≥2. With respect to the Sum of Arithmetic Sequence, a + c = c( 2s+c-1)2. Proposed contradiction: Let c( 2s +c-1)2 be a power of 2. Then c and 2s + c - 1 are powers of 2. Since c≥ 2, c is even. Hence, 2s + c - 1is odd. Since 2s + c - 1 is odd and a power of 2, 2s +c-1=1 However, 2s + c +1 > 2(0) +2 -1 = 1. This is a contradiction. Then, any sum of two or more consecutive integers cannot be 2n.
Transcribed Image Text:Theorem: A positive integer n will be the sum of a sequence of consecutive positive integers if and only if n #2^n. Proof: Let s be the smallest of c consecutive integers € Z + such that c≥2. With respect to the Sum of Arithmetic Sequence, a + c = c( 2s+c-1)2. Proposed contradiction: Let c( 2s +c-1)2 be a power of 2. Then c and 2s + c - 1 are powers of 2. Since c≥ 2, c is even. Hence, 2s + c - 1is odd. Since 2s + c - 1 is odd and a power of 2, 2s +c-1=1 However, 2s + c +1 > 2(0) +2 -1 = 1. This is a contradiction. Then, any sum of two or more consecutive integers cannot be 2n.
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