Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX E - Sigma Notation - E Exercises: 45

Answer

$14$

Work Step by Step

Find the limit $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{2}{n}[(\frac{2i}{n})^{3}+5(\frac{2i}{n})]$ $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{2}{n}[(\frac{2i}{n})^{3}+5(\frac{2i}{n})]=\lim\limits_{n \to \infty}(\frac{16}{n^{4}}\sum\limits_{i =1}^{n}{i}^{3}+\frac{20}{n^{2}}\sum\limits_{i =1}^{n}i)$ $ \sum \limits_{i =1}^{n}i^{3}=[\frac{n(n+1)}{2}]^{2}$ Thus, $=\lim\limits_{n \to \infty}[\frac{16}{n^{4}}(\frac{n(n+1)}{2})^{2}+\frac{20}{n^{2}}(\frac{n(n+1)}{2})]$ $=\lim\limits_{n \to \infty}[4+\frac{8}{n}+\frac{4}{n^{2}}+10+\frac{10}{n}]$ $=4+0+0+10+0$ Hence, $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{2}{n}[(\frac{2i}{n})^{3}+5(\frac{2i}{n})]=14$
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