Answer
$=\Sigma^{k}_{k=1}i(i!)+(k+1)(k+1)!$
Work Step by Step
By separating of the final term
$\Sigma^{k+1}_{j=1}i(i!)=\Sigma^{k}_{j=1}i(i!)+(k+1)(k+1)!$
Discrete Mathematics with Applications 4th Edition
by Epp, Susanna S.
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: 37
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