Question

The sum of series $4-9x+16x^2-25x^3+36x^4-49x^5+\dots +\infty$ is

• A. $\frac{4+3x+x^2}{(1+x{)}^3}$
• B. $\frac{x^2-3x-4}{(1+x{)}^3}$
• C. $\frac{x^2+3x-4}{(1+x{)}^3}$
• D. $\frac{x^2-3x+4}{(1+x{)}^3}$

Answer 1

The correct option is A $\frac{4+3x+x^2}{(1+x{)}^3}$

Let
$S=4-9x+16x^2-25x^3+36x^4-49x^5+\dots \infty$
$xS=4x-9x^2+16x^3-25x^4+36x^5-\dots \infty$

On adding, we get
$S(1+x)=4-5x+7x^2-9x^3+11x^4-13x^5+\dots \infty$
$=4-x[5-7x+9x^2-11x^3+...\infty ]$

Since, $5-7x+9x^2-11x^3+...\infty$ is AGP with $a=5,d=2,r=-x$
$\Rightarrow S(1+x)=4-x[\frac{5x}{1+x}+\frac{2(-x)}{(1+x{)}^2}]$
$\Rightarrow S=\frac{4+3x+x^2}{(1+x{)}^3}$


Alternate Solution:
Let
$S=4-9x+16x^2-25x^3+36x^4-49x^5+\dots \infty$
$xS=4x-9x^2+16x^3-25x^4+36x^5-\dots \infty$

On adding, we get
$S(1+x)=4-5x+7x^2-9x^3+11x^4-13x^5+\dots \infty$
$S(1+x)x=4x-5x^2+7x^3-9x^4+11x^5-\dots \infty$

On adding, we get
$S(1+x{)}^2=4-x+2x^2-2x^3+2x^4-2x^5+\dots \infty$
$\Rightarrow S(1+x{)}^2=4-x+2x^2(1-x+x^2-\dots \infty )$
$\Rightarrow S(1+x{)}^2=4-x+\frac{2x^2}{1+x}=\frac{4+3x+x^2}{1+x}$
$\Rightarrow S=\frac{4+3x+x^2}{(1+x{)}^3}$