Question

Find the $n^{th}$ term and the sum to $n$ terms of the following series: $2+7x+25x^2+91x^3+....$

Answer 1

Given Series $S=2+7x+25x^2+91x^3+....$

$a_1=2=(3^0+4^0)a_3=(3^2+4^2)x^2{a}_2=7x=(3^1+4^1)x{a}_4=(3^3+4^3)x^3$

Hence,

$a_n=(3^{n-1}+4^{n-1})x^{n-1}$

Let the scale of relation be $1-px-q{x}^2$

Taking coefficient of $x^0,x^1,x^2$ we get $25-7p-2q=0\left(1\right)$

Taking coefficient of $x^1,x^2,x^3$ we get $91-25p-7q=0\left(2\right)$

$\left(1\right)\times 7-\left(2\right)\times 2$ we get

$175=49p+14q-182=-50p+(-14q)-7=-pp=7$

Putting this value of $p$ in equation $\left(1\right)$

$25-7\times 7=2q-24=2qq=-12$

Scale of relation is $1-7x+12x^2$

$S=2+7x+25x^2+91x^3+....-7xS=-14x-49x^2-175x^3-...12x^{2}S=24x^2+84x^3$

Adding above equations,

$(1-7x+12x^2)S=2-7xS=\frac{2-7x}{1-7x+12x^2}$