Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 5 - Sequences, Mathematical Induction, and Recursion - Exercise Set 5.3 - Page 266: 4

Answer

$1-4+9-16+\cdots+(-1)^{n+1}n^2=(-1)^{n+1}(1+2+3+4+\cdots+n)=(-1)^{n+1}(\frac{n(n+1)}{2})$

Work Step by Step

Based on the existing pattern, we can write a general formula as: $1-4+9-16+\cdots+(-1)^{n+1}n^2=(-1)^{n+1}(1+2+3+4+\cdots+n)=(-1)^{n+1}(\frac{n(n+1)}{2})$ Proof: Let $P(n)$ be the above statement. 1. For $n=1$, we have $LHS=(-1)^21=1$ and $RHS=(-1)^2(\frac{1(1+1}{2})=1=LHS$, thus it is true for $P(1)$. 2. Assume it is true for $P(k)$, that is: $1-4+9-16+\cdots+(-1)^{k+1}k^2=(-1)^{k+1}(1+2+3+4+\cdots+k)=(-1)^{k+1}(\frac{k(k+1)}{2})$ 3. For $n=k+1$, we have: $LHS=1-4+9-16+\cdots+(-1)^{k+1}k^2+(-1)^{k+2}(k+1)^2\\ =(-1)^{k+1}(1+2+3+4+\cdots+k)+(-1)^{k+2}(k+1)^2\\ =(-1)^{k+1}(\frac{k(k+1)}{2})+(-1)^{k+2}(k+1)^2\\ =(-1)^{k+2}(2k^2+4k+2-k^2-k)\frac{1}{2} =(-1)^{k+2}(\frac{k^2+3k+2}{2}) =(-1)^{k+2}\frac{(k+1)(k+2)}{2}=RHS$ 4. Thus $P(k+1)$ is also true and we proved the statement by mathematical induction.
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