Chapter 12.3, Problem 41AYU

In problems 41-46, find each sum.

$\frac{1}{4}+\frac{2}{4}+\frac{2^2}{4}+\frac{2^3}{4}+\cdots +\frac{2^{n-1}}{4}$

To determine

To calculate: To find the sum of $n$ terms of the given geometric series $\frac{1}{4}+\frac{2}{4}+\frac{2^2}{4}+\frac{2^3}{4}+\cdots +\frac{2^{n-1}}{4}$.

Answer to Problem 41AYU

Solution:

The sum of $n$ terms of the given geometric series $S_n=-\frac{1}{4}\cdot (1-2^n)$

Explanation of Solution

Given:

Series is given as follows:

$\frac{1}{4}+\frac{2}{4}+\frac{2^2}{4}+\frac{2^3}{4}+\cdots +\frac{2^{n-1}}{4}=\frac{1}{4}\left(1+2+2^2+2^3+\cdots +2^{n-1}\right)$

Formula used:

The sum of $n$ terms of the geometric sequence is given by

$S_n=a_1\frac{1-r^n}{1-r}$

Consider the series, the first term value is 1 and the common ratio $=r=\frac{a_n}{a_{n-1}}=2$.

Sum of $n$ terms of the series $S_n=a_{1 }\frac{1-r^n}{1-r}=\frac{1}{4}\left\{1\cdot \frac{1-2^n}{1-2}\right\}=-\frac{1}{4}\left(1-2^n\right)$.