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Q 60.

Expert-verified
Found in: Page 809

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# In Problems 51–60, write out each sum. $\sum _{k=3}^{n}{\left(-1\right)}^{k+1}{2}^{k}$

The required sum is $\sum _{k=3}^{n}{\left(-1\right)}^{k+1}{2}^{k}==\left\{8-16+32-........+{\left(-1\right)}^{n+1}·{2}^{n}\right\}$.

See the step by step solution

## Step 1. Write the given information.

The given summation notation is:

$\sum _{k=3}^{n}{\left(-1\right)}^{k+1}{2}^{k}$

## Step 2. Write out the sum.

To get the sum, we need to substitute $3,4,5,......,n$ for k, in the given sequence $\sum _{k=3}^{n}{\left(-1\right)}^{k+1}{2}^{k}$,

$\sum _{k=3}^{n}{\left(-1\right)}^{k+1}{2}^{k}=\left\{{\left(-1\right)}^{3+1}{2}^{3}+{\left(-1\right)}^{4+1}{2}^{4}+{\left(-1\right)}^{5+1}{2}^{5}+......+{\left(-1\right)}^{n+1}{2}^{n}\right\}\phantom{\rule{0ex}{0ex}}=\left\{{\left(-1\right)}^{4}·8+{\left(-1\right)}^{5}·16+{\left(-1\right)}^{6}·32+.......+{\left(-1\right)}^{n+1}·{2}^{n}\right\}\phantom{\rule{0ex}{0ex}}=\left\{8-16+32-........+{\left(-1\right)}^{n+1}·{2}^{n}\right\}$

Therefore, localid="1646756404405" $\sum _{k=3}^{n}{\left(-1\right)}^{k+1}{2}^{k}==\left\{8-16+32-........+{\left(-1\right)}^{n+1}·{2}^{n}\right\}$.