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 5.1 - Page 243: 55

Answer

$\Sigma^{n+1}_{i=0}\frac{(i-1)^2}{i.2}$ =$\Sigma^{n}_{j=1}\frac{j^2}{jn+n}$

Work Step by Step

When i = 1, then j = 0. When i = n + 1, then j = n. Since j = i − 1, then i = j + 1. Thus, $\frac{(i-1)^2}{i.n}=\frac{((j+1)-1)^2}{(j+1).n}=\frac{j^2}{jn+n}$ N is constant so $\Sigma^{n+1}_{i=0}\frac{(i-1)^2}{i.2}$ =$\Sigma^{n}_{j=1}\frac{j^2}{jn+n}$
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