Question

Sum to $infinite$ terms the following series:
$1+5x^2+9x^4+13x^6+....,|x|<1$

Answer 1

$S=1+5x^2+9x^4+13x^6+...$ $............\left(1\right)$

$S{x}^2=x^2+5x^4+9x^6+....$ $...............\left(2\right)$

$e{q}^n\left(1\right)–e{q}^n\left(2\right)$, we get

$(1–x^2)S=1+4x^2+4x^4+4x^6+\dots ...$ $.........\left(3\right)$

This is a $G.P$. of common ratio $=x^2$

So, using sum of $G.P.=\frac{a}{(1-r)}$

Here, $a=firstterm$ and $r=commonratio$

So,

$(1–x^2)S=1+\frac{4x^2}{(1–x^2)}$

$(1–x^2)S=\frac{1-x^2+4x^2}{(1–x^2)}$

$S=\frac{1+3x^2}{(1–x^2{)}^2}$

Hence, this is the answer.