Chapter F, Problem 49E

To determine

To evaluate: The sum ${\displaystyle \sum _{i=1}^n\left(2i+2^i\right)}$.

Answer to Problem 49E

The value of the sum ${\displaystyle \sum _{i=1}^n\left(2i+2^i\right)}$ is $2^{n+1}+n^2+n-2$.

Explanation of Solution

The formula for the sum of a finite geometric serious with first term a and common ratio $r\ne 1$, is ${\displaystyle \sum _{i=1}^{n}a{r}^{i-1}=a+ar+a{r}^2+\cdots +a{r}^{n-1}}=\frac{a\left(r^n-1\right)}{r-1}$.

The sum ${\displaystyle \sum _{i=1}^n\left(2i+2^i\right)}$ is simplified as ${\displaystyle \sum _{i=1}^{n}2i}+{\displaystyle \sum _{i=1}^n{22}^{i-1}}$.

Here, the sum ${\displaystyle \sum _{i=1}^n{22}^{i-1}}$ is in the form ${\displaystyle \sum _{i=1}^{n}a{r}^{i-1}}$ where, $a=2\text{and}r=2$.

By the above formula, the value of the sum ${\displaystyle \sum _{i=1}^n{22}^{i-1}}$ is simplified as,

$\begin{array}{c}{\displaystyle \sum _{i=1}^n{22}^{i-1}}=\frac{2\left[2^n-1\right]}{2-1}\\ =2\left(2^n-1\right)\end{array}$

Thus, the value of the sum ${\displaystyle \sum _{i=1}^n{22}^{i-1}}$ is $2\left(2^n-1\right)$.

Simplify the expression ${\displaystyle \sum _{i=1}^n\left(2i+2^i\right)}$ and obtain the value of the sum.

$\begin{array}{c}{\displaystyle \sum _{i=1}^n\left(2i+2^i\right)}={\displaystyle \sum _{i=1}^{n}2i}+{\displaystyle \sum _{i=1}^n{22}^{i-1}}\\ =2\frac{n\left(n+1\right)}{2}+2\left(2^n-1\right)\\ =n\left(n+1\right)+2\left(2^n-1\right)\\ =2^{n+1}+n^2+n-2\end{array}$

Thus, the value of the sum ${\displaystyle \sum _{i=1}^n\left(2i+2^i\right)}$ is $2^{n+1}+n^2+n-2$.