Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Appendix E - Sigma Notation - E Exercises - Page A38: 29



Work Step by Step

We simplify: $\displaystyle \sum_{i=1}^{n}2i$ We can take out $2$ because it is a constant $(\sum{cx}=c\sum{x})$: $=2\sum_{i=1}^{n}i$ We know that $\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$, so: $=2*\frac{n(n+1)}{2}$ $=n(n+1)$
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