Chapter 8, Problem 13CT

To determine

To find sum of the below sequence−

  ${{\displaystyle \sum _{n=1}^{8}49\left(\frac{1}{7}\right)}}^{n-1}$

Answer to Problem 13CT

Thesum is $\frac{960800}{16897}$ .

Explanation of Solution

Given: Sequence: ${{\displaystyle \sum _{n=1}^{8}49\left(\frac{1}{7}\right)}}^{n-1}$

Calculation:

Given: Sequence: ${{\displaystyle \sum _{n=1}^{8}49\left(\frac{1}{7}\right)}}^{n-1}$

Sequence can be written as:

  $\begin{array}{l}{{\displaystyle \sum _{n=1}^{8}49\left(\frac{1}{7}\right)}}^{n-1}={\displaystyle \sum _{n=1}^{8}49{\left({\left(7\right)}^{-1}\right)}^{n-1}}\\ {{\displaystyle \sum _{n=1}^{8}49\left(\frac{1}{7}\right)}}^{n-1}={\displaystyle \sum _{n=1}^8{7}^2{\left(7\right)}^{-n+1}}\\ {{\displaystyle \sum _{n=1}^{8}49\left(\frac{1}{7}\right)}}^{n-1}={\displaystyle \sum _{n=1}^8{7}^{2-n+1}}\\ {{\displaystyle \sum _{n=1}^{8}49\left(\frac{1}{7}\right)}}^{n-1}={\displaystyle \sum _{n=1}^8{7}^{\left(3-n\right)}}\end{array}$

Sum $=$ ${\displaystyle \sum _{n=1}^8{7}^{\left(3-n\right)}}$

  $\begin{array}{l}{\displaystyle \sum _{n=1}^8{7}^{\left(3-n\right)}}=7^2+7^1+7^0+7^{-1}+7^{-2}+7^{-3}+7^{-4}+7^{-5}\\ {\displaystyle \sum _{n=1}^8{7}^{\left(3-n\right)}}=\frac{960800}{16897}\end{array}$

Conclusion:

Hence, the sum is $\frac{960800}{16897}$ .