Question

The value of the $\sum _{n=0}^{\infty }\frac{2n+3}{3^n}$ is equal to

Answer 1

We have
$S=\sum _{n=0}^{\infty }\frac{2n+3}{3^n}=\underset{S_1}{\underset{}{\sum _{n=0}^{\infty }\frac{3}{3^n}}}+\underset{S_2}{\underset{}{\sum _{n=0}^{\infty }\frac{2n}{3^n}}}$

Now, $S_1=\sum _{n=0}^{\infty }\frac{3}{3^n}=3+1+\frac{1}{3}+\frac{1}{3^2}+\dots \infty$
$\therefore S_1=3+\frac{1}{1-\left(\frac{1}{3}\right)}=\frac{9}{2}$
$S_2=\sum _{n=0}^{\infty }\frac{2n}{3^n}$
$\Rightarrow S_2=\frac{2}{3}+\frac{4}{3^2}+\frac{6}{3^3}+\frac{8}{3^4}\dots \infty$
$\Rightarrow \frac{S_2}{3}=\frac{2}{3^2}+\frac{4}{3^3}+\frac{6}{3^4}+\dots \infty$

On subtracting, we get
$\frac{2S_2}{3}=\frac{2}{3}[1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\dots \infty ]$
$\therefore S_2=\frac{1}{1-\left(\frac{1}{3}\right)}=\frac{3}{2}$
Hence, $S=\frac{9}{2}+\frac{3}{2}=6$