Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 11 - Sequences; Induction; the Binomial Theorem - Chapter Test - Page 860: 5


$\sum_{n=1}^{10} (-1)^n(\dfrac{n+1}{n+4})$

Work Step by Step

We are given the sum: $-\dfrac{2}{5}+\dfrac{3}{6}-\dfrac{4}{7}+...+\dfrac{11}{14}$ We notice that the numerator shows the numbers from $2$ to $11$, increasing by $1$ each time. Thus, we have $11-2+1=10$ terms in total and each term can be represented in the numerator as $n+1$. We also notice that the denominator is always greater than the numerator by $3$. Thus, we can represent the denominator as $n+1+3=n+4$. Finally, we notice that the terms alternate in sign, so we multiply each term with $(-1)^n$ . Therefore, the required summation formula is: $\sum_{n=1}^{10} (-1)^n(\dfrac{n+1}{n+4})$
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