Question

In Problems 61–70, express each sum using summation notation.

$3+\frac{3^2}{2}+\frac{3^3}{3}+.....+\frac{3^n}{n}$

Short answer

The required summation notation is $3+\frac{3^2}{2}+\frac{3^3}{3}+.....+\frac{3^n}{n}=\sum _{k=1}^n\left(\frac{3^k}{k}\right)$.

Step-by-step solution

Step 1. Write the given information.

The given sum is:

$3+\frac{3^2}{2}+\frac{3^3}{3}+.....+\frac{3^n}{n}$

Step 2. Write the sum using summation notation.

As we can see that the sum $3+\frac{3^2}{2}+\frac{3^3}{3}+.....+\frac{3^n}{n}$has n terms, each term represents the form $\frac{3^k}{k}$, and starts at $k=1$and ends at $k=n$.

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

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