Chapter 13.1, Problem 64SR

To determine

To find : Sum of geometric series.

Answer to Problem 64SR

  $32552$

Explanation of Solution

Given information :

We know that, the given value is ${\displaystyle \sum _{k=1}^8\frac{1}{3}\cdot 5^{R-1}}$

Calculation :

The given equation is

  ${\displaystyle \sum _{k=1}^8\frac{1}{3}\cdot 5^{R-1}=\frac{1}{3}(1+5+5^2+5^3+\mathrm{...}+5^7)}$

  $\therefore$ Common ratio = $5$ and a = $1$

As $S_n=\frac{a(r^n-1)}{r-1},r>1$

  $\begin{array}{l}{S}_8=\frac{1(5^8-1)}{5-1}=\frac{5^8-1}{4}\\ \Rightarrow {\displaystyle \sum _{k=1}^8\frac{1}{3}\cdot 5^{k-1}=\frac{1}{3}\frac{(5^8-1)}{4}}\\ =\frac{1}{3}\frac{(390624)}{4}\\ =\frac{97656}{3}\\ =32552\end{array}$