Question

The sum of the infinite series $\sum _{n=1}^{\infty }(\frac{n^2+3}{2^n})$ is equal to

Answer 1

$\sum _{n=1}^{\infty }\frac{n^2}{2^n}+\sum _{n=1}^{\infty }\frac{3}{2^n}$
Let $x=\sum _{n=1}^{\infty }\frac{n^2}{2^n}$ & $y=\sum _{n=1}^{\infty }\frac{3}{2^n}$
$x=\frac{1}{2}+\frac{4}{2^2}+\frac{9}{2^3}+\frac{16}{2^4}+\frac{25}{2^5}+......\infty$ (1)
Subtracting (2) from (1)
$\frac{x}{2}=\frac{1}{2^2}+\frac{4}{2^3}+\frac{9}{2^4}+\frac{16}{2^5}+......\infty$ (2)
$\frac{x}{2}=\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+\frac{9}{2^5}+......\infty$ (3)
$\frac{x}{4}=\frac{1}{2^2}+\frac{3}{2^3}+\frac{5}{2^4}+\frac{7}{2^5}+......\infty$ (4)

Subtract (4) from (3),
$\frac{x}{4}=\frac{1}{2}+2[\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+......\infty ]$
$\frac{x}{4}=\frac{1}{2}+2\left(\frac{1/4}{1-1/2}\right)=\frac{1}{2}+2.\frac{1}{2}=\frac{3}{2}$
$\Rightarrow x=6$ and $y=\frac{3}{2}+\frac{3}{2^2}+\frac{3}{2^3}+\frac{3}{2^4}+.......\infty$
$=(3\frac{1/2}{1-1/2})=3$
$\therefore sum=x+y=6+3=9$