Chapter 14.1, Problem 62SR

To determine

To find :Sum of given geometric series.

Answer to Problem 62SR

  $\frac{31}{4}$

Explanation of Solution

Given information :

We know that, the given value is ${\displaystyle \sum _{k=1}^5\frac{1}{4}\cdot 2^{k-1}}$

Calculation :

The given equation is

  $\frac{1}{4}(1+2+2^2+2^3+2^4)$

Common ratio (r) = $2$ and a = $1$

  $\begin{array}{l}{S}_n=\frac{a(r^n-1)}{r-1},r>1\\ S_5=\frac{1(2^5-1)}{2-1}=\frac{32-1}{1}=31\\ \text{So},{\displaystyle \sum _{k=1}^5\frac{1}{4}\cdot 2^{k-1}=\frac{1}{4}{\displaystyle \sum _{k=1}^5{2}^{k-1}=\frac{1}{4}\times 31}}\\ =\frac{31}{4}\end{array}$