## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# APPENDIX E - Sigma Notation - E Exercises: 30

#### Answer

$-\frac{n(5n+1)}{2}$

#### Work Step by Step

We simplify: $\displaystyle \sum_{i=1}^{n}(2-5i)$ We know that we can split up the sums ($\sum{(x+y)}=\sum{x}+\sum{y}$): $=\sum_{i=1}^{n}2-\sum_{i=1}^{n}5i$ We can take out $5$ because it is a constant $(\sum{cx}=c\sum{x})$: $=2n-5\sum_{i=1}^{n}i$ We know that $\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$, so: $=2n-\frac{5n(n+1)}{2}$ $=\frac{4n}{2}-\frac{5n^{2}+5n}{2}$ $=-\frac{4n-5n^2-5n}{2}$ $=-\frac{-n-5n^2}{2}$ $=-\frac{n(5n+1)}{2}$

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