## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 5 - Section 5.2 - Sigma Notation and Limits of Finite Sums - Exercises - Page 299: 12

#### Answer

$\sum\limits_{k=1}^4k^2$

#### Work Step by Step

$1=1^2, 4=2^2, 9=3^2, 16=4^2$, therefore the additives are the first four square numbers, therefore the formula is: $\sum\limits_{k=1}^4k^2$

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