Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Practice Exercises - Page 307: 3

Answer

a. $-\frac{1}{2}$ b. $31$ c. $13$ d. $0$

Work Step by Step

Given $\Sigma_{k=1}^{10} a_k=-2$ and $\Sigma_{k=1}^{10} b_k=25$, we have: a. $\Sigma_{k=1}^{10} \frac{a_k}{4}=\frac{1}{4}\Sigma_{k=1}^{10} a_k=-\frac{1}{2}$ b. $\Sigma_{k=1}^{10} (b_k-3a_k)=\Sigma_{k=1}^{10} b_k-3\Sigma_{k=1}^{10} a_k=25-3(-2)=31$ c. $\Sigma_{k=1}^{10} (a_k+b_k-1)=\Sigma_{k=1}^{10} a_k+\Sigma_{k=1}^{10} b_k-\Sigma_{k=1}^{10} 1=(-2)+25-10=13$ d. $\Sigma_{k=1}^{10} (\frac{5}{2}-b_k)=\Sigma_{k=1}^{10} \frac{5}{2}-\Sigma_{k=1}^{10} b_k=\frac{5}{2}\times10-25=0$
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