Question

In Problems 51–60, write out each sum.

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

Short answer

The required sum is$\sum _{k=2}^n{\left(-1\right)}^k\mathrm{In}k=\left\{\mathrm{In}2-\mathrm{In}3+\mathrm{In}4-.......+{\left(-1\right)}^{\mathrm{n}}\mathrm{In}n\right\}$.

Step-by-step solution

Step 1. Write the given information.

The given summation notation is:

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

Step 2. Write out the sum.

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

$$\begin{gathered} \sum _{k=2}^n{\left(-1\right)}^k\mathrm{In}k=\left\{{\left(-1\right)}^2\mathrm{In}2+{\left(-1\right)}^3\mathrm{In}3+{\left(-1\right)}^4\mathrm{In}4+.......+{\left(-1\right)}^{\mathrm{n}}\mathrm{In}n\right\} \\ =\left\{\mathrm{In}2-\mathrm{In}3+\mathrm{In}4-.......+{\left(-1\right)}^{\mathrm{n}}\mathrm{In}n\right\} \end{gathered}$$

Therefore,$\sum _{k=2}^n{\left(-1\right)}^k\mathrm{In}k=\left\{\mathrm{In}2-\mathrm{In}3+\mathrm{In}4-.......+{\left(-1\right)}^{\mathrm{n}}\mathrm{In}n\right\}$.