Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Cumulative Review Exercises - Page 1128: 35


The value of $\sum\limits_{i=1}^{50}{\left( 4i-25 \right)}$ is $3850$. We have to find the value of the given summation from $ i=1$ to $ i=50$.

Work Step by Step

$\begin{align} & \sum\limits_{i=1}^{50}{\left( 4i-25 \right)}=\left[ \left( 4\cdot 1-25 \right)+\left( 4\cdot 2-25 \right)+\left( 4\cdot 3-25 \right)+\ldots +\left( 4\cdot 50-25 \right) \right] \\ & =\left[ 4\cdot \left( 1+2+3+\ldots +50 \right)-25-25-\ldots -25 \right] \end{align}$ We have taken 4 as a common number from the first digit of every term and obtained an arithmetic sequence. We expand this in the form: $\begin{align} & \left[ \frac{4\left( n \right)\left( n+1 \right)}{2}-25n \right]=\left[ \frac{4\left( 50 \right)\left( 51 \right)}{2}-25\left( 50 \right) \right] \\ & =\left[ 5100-1250 \right] \\ & =3850 \end{align}$ Hence, the value of the sum $\sum\limits_{i=1}^{50}{\left( 4i-25 \right)}$ is $3850$.
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