Question

Find the sum of the geometric series $4+2+1+...+$ $\frac{1}{16}$

• A. $\frac{17}{16}$
• B. $\frac{107}{16}$
• C. $\frac{117}{16}$
• D. $\frac{127}{16}$

Answer 1

The correct option is D $\frac{127}{16}$

Given series is $4+2+1+.....+\frac{1}{16}$

First term, $a$ is $4$
Common ratio, $r=$ $\frac{2}{4}=\frac{1}{2}$
Use the formula for the sum of the geometric series.
$a{r}^n$ is a next term of $\frac{1}{16}$
$\therefore a_{n+1}=a{r}^n=\frac{1}{2}\times \frac{1}{16}=\frac{1}{32}$ is the next term.
$S=\frac{a-a{r}^{n+1}}{1-r}$
$S=\frac{4-\frac{1}{32}}{1-\frac{1}{2}}$
$S=\frac{\frac{127}{32}}{\frac{1}{2}}$
$S=\frac{127}{16}$