Answer
$30$
Work Step by Step
$\sum ^{4}_{k=1}k^{2}=\dfrac {n\left( n+1\right) \left( 2n+1\right) }{6}=\dfrac {4\times \left( 4+1\right) \times \left( 2\times 4+1\right) }{6}=30$
Precalculus: Mathematics for Calculus, 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305071751
ISBN 13:
978-1-30507-175-9
Chapter 12 - Section 12.1 - Sequences and Summation Notation - 12.1 Exercises - Page 851: 48
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