Chapter 14.7, Problem 8MRE

To determine

To write: The series using sigma notation and then find the sum.

Answer to Problem 8MRE

The given series in sigma notation is ${\displaystyle \sum _{n=1}^{n}100{\left(\frac{1}{5}\right)}^{n-1}}$ and its sum is $125\left(1-{\left(\frac{1}{5}\right)}^n\right)$ .

Explanation of Solution

Given: $100+20+4+\mathrm{....}$ .

Consider the given series.

  $3+12+48+\mathrm{...}+3072$

The given series is geometric with common ratio $\frac{1}{5}$ .

The $n^{\text{th}}$ term of the GP is given as,

  $t_n=a_1\cdot r^{n-1}$

Substitute the value in the above expression.

  $t_n=100{\left(\frac{1}{5}\right)}^{n-1}$

The series in sigma notation is written as,

  ${\displaystyle \sum _{n=1}^{n}100{\left(\frac{1}{5}\right)}^{n-1}}$

The sum of the series is calculated as,

  $\begin{array}{c}{S}_n=\frac{a_1\left(1-r^n\right)}{1-r}\\ S_n=\frac{100\left(1-{\frac{1}{5}}^n\right)}{1-\frac{1}{5}}\\ =125\left(1-{\left(\frac{1}{5}\right)}^n\right)\end{array}$

Therefore, the given series in sigma notation is ${\displaystyle \sum _{n=1}^{n}100{\left(\frac{1}{5}\right)}^{n-1}}$ and its sum is $125\left(1-{\left(\frac{1}{5}\right)}^n\right)$ .