## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 5: Integrals - Section 5.2 - Sigma Notation and Limits of Finite Sums - Exercises 5.2 - Page 265: 24

61

#### Work Step by Step

6$\Sigma _{k=1}$(k^2-5) =6$\Sigma _{k=1} k^2$- $6\Sigma _{k=1}$ 5 =$\frac{6(6+1)(2(6)+1)}{6} - 5(6)$ =61

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