Chapter 12.3, Problem 39AYU

To determine

To find: The sum of series.

Answer to Problem 39AYU

The sum of series is $S_n=-\frac{1}{4}\left(1-2^n\right)$ .

Explanation of Solution

Given information:

The series is $\frac{1}{4}+\frac{2}{4}+\frac{2^2}{4}+\frac{2^3}{4}+\mathrm{............}+\frac{2^{n-1}}{4}$ .

Calculation:

For the given series $a_1=\frac{1}{4}$ , $a_2=\frac{2}{4}$ and $a_3=\frac{2^2}{4}$ .

Calculate $\frac{a_2}{a_1}$ .

  $\begin{array}{c}\frac{a_2}{a_1}=\frac{\frac{2}{4}}{\frac{1}{4}}\\ =2\end{array}$

Calculate $\frac{a_3}{a_1}$ .

  $\begin{array}{c}\frac{a_3}{a_1}=\frac{\frac{2^2}{4}}{\frac{2}{4}}\\ =2\end{array}$

Since the ratio of successive terms is same it is a geometric series with common ratio $r=2$ .

The sum of first $n$ terms of a geometric sequence is calculated by,

  $S_n=a_1\frac{1-r^n}{1-r}$

Substitute the given values.

  $\begin{array}{l}{S}_n=\frac{1}{4}\left(\frac{1-2^n}{1-2}\right)\\ S_n=\frac{1}{4}\frac{\left(1-2^n\right)}{-1}\\ S_n=-\frac{1}{4}\left(1-2^n\right)\end{array}$

Therefore, the sum of series is $S_n=-\frac{1}{4}\left(1-2^n\right)$ .