Question

In Problems 51–60, write out each sum.

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

Short answer

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

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

$$\begin{gathered} \sum _{k=3}^n{\left(-1\right)}^{k+1}{2}^k=\left\{{\left(-1\right)}^{3+1}{2}^3+{\left(-1\right)}^{4+1}{2}^4+{\left(-1\right)}^{5+1}{2}^5+......+{\left(-1\right)}^{n+1}{2}^n\right\} \\ =\left\{{\left(-1\right)}^4·8+{\left(-1\right)}^5·16+{\left(-1\right)}^6·32+.......+{\left(-1\right)}^{n+1}·2^n\right\} \\ =\left\{8-16+32-........+{\left(-1\right)}^{n+1}·2^n\right\} \end{gathered}$$

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