Question

If $S=1+2^{2}x+3^2{x}^2+4^2{x}^3+......\infty$ where $x=\frac{1}{5}$. Find the value of 32 S.

Answer 1

This series is not AP, GP, HP and AGP

How can we proceed?

But we see $1^2,2^2,3^2,4^2..........\infty$ are square numbers and $1,x,x^2,x^3,x^4,........\infty$ is a geometric progression.
If we subtract square numbers. We can form an AP.

To subtract square numbers multiply given series by common ratio and then subtract it with original series.

$s=1+2^{2}x+3^2{x}^2+4^2{x}^3+......\infty$ -----(1)
Multiply equation 1 by common ratio x
$xs=x+2^2{x}^2+3^2{x}^3+.....\infty$ -----(2)
Subtract equation 2 from equation 1
$(1-x)s=1+3x+5x^2+7x^3+....\infty$ -----(3)
Now, we will see we got Arithmetic - geometric progression.
At this stage you should be able to solve a AGP
Again multiply equation 3 by common ratio x
$x(1-x)s=x+3x^2+5x^2+7x^4+....\infty$ ----(4)
Subtracting equation 4 from equation 3, we get
$(1-x)(1-x)s=1+2(x+x^2+x^3+......\infty )$
$(1-x{)}^{2}s=1+2(\frac{x}{1+x})=\frac{1+x}{1-x}$

$s=\frac{1+x}{(1-x{)}^3}$
Substitute x =$\frac{1}{5}$
$s=\frac{1+\frac{1}{5}}{(1-\frac{1}{5}{)}^3}=\frac{75}{32}$
32 x 5 = 75