Question

Find sum to n terms of the series $3+5\times \frac{1}{4}+7\times \frac{1}{4^2}+$_____.

Answer 1

Given:

$3+5\times \frac{1}{4}+7\times \frac{1}{4^2}+\cdots$

Let $S=3+5\times \frac{1}{4}+7\times \frac{1}{4^2}+\cdots$. …… (1)

Multiply both sides by $\frac{1}{4}$.

$\frac{1}{4}S=\frac{3}{4}+5\times \frac{1}{4^2}+7\times \frac{1}{4^3}+\cdots$ …… (2)

Subtract equation (2) from equation (1).

$S-\frac{1}{4}S=3+(\frac{5}{4}-\frac{3}{4})+(7\times \frac{1}{4^2}-5\times \frac{1}{4^2})+(9\times \frac{1}{4^3}-7\times \frac{1}{4^3})+\cdots$

$\frac{3}{4}S=3+\frac{2}{4}+\frac{2}{4^2}+\frac{2}{4^3}+\cdots$

$\frac{3}{4}S=3+\frac{1}{2}(1+\frac{1}{4}+\frac{1}{4^2}+\cdots +(n–1)terms)$

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

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

$S=4-\frac{8}{9}({\left(\frac{1}{4}\right)}^{n-1}-1)$

Thus, the sum of n terms of the series is $S=4-\frac{8}{9}({\left(\frac{1}{4}\right)}^{n-1}-1)$.