If ( |a|<1, ) then ( 1+2a+3{{a}^{2}}+4{{a}^{3}}+..... ) is e... | Mathematics | SolveeIt

Question

If $|a|<1,$ then $1+2a+3{{a}^{2}}+4{{a}^{3}}+.....$ is equal to

$\frac{1}{1-a}$

$\frac{1}{1+a}$

$\frac{1}{1+{{a}^{2}}}$

$\frac{1}{{{(1-a)}^{2}}}$

Answer

$\frac{1}{{{(1-a)}^{2}}}$

Explanation

Solution

The correct answer is(D): $=\frac{1}{{{(1-a)}^{2}}}.$

Given series, $1+2a+3{{a}^{2}}+4{{a}^{3}}+....$
is a arithmetic geometric series. Here,
${{a}_{1}}=1,\,\,d=1,\,\,r=a$
$\therefore$ ${{S}_{\infty }}=\frac{{{a}_{1}}}{1-r}+\frac{d.r}{{{(1-r)}^{2}}}$
$=\frac{1}{1-a}+\frac{1.a}{{{(1-a)}^{2}}}$
$=\frac{1}{{{(1-a)}^{2}}}$

Mathematics Question on Sequence and series

Solution