Question

Evaluate $\frac{x(x+y)+x^2(x^2+y^2)+x^3(x^3+y^3)+....tonterms}{x^2\frac{(1-x^{2n})}{1-x^2}+xy\frac{(1-x^n{y}^n)}{1-xy}}$

Answer 1

We have,

$\frac{x(x+y)+x^2(x^2+y^2)+x^3(x^3+y^3)+....tonterms}{x^2\frac{(1-x^{2n})}{1-x^2}+xy\frac{(1-x^n{y}^n)}{1-xy}}$

$\Rightarrow \frac{x^2+xy+x^4+x^2{y}^2+x^6+x^3{y}^3+....tonterms}{x^2\frac{(1-x^{2n})}{1-x^2}+xy\frac{(1-x^n{y}^n)}{1-xy}}$

$\Rightarrow \frac{(x^2+x^4+x^6+......+tonterms)+(xy+x^2{y}^2+x^3{y}^3+....tonterms)}{x^2\frac{(1-x^{2n})}{1-x^2}+xy\frac{(1-x^n{y}^n)}{1-xy}}$

We know that sum of $n$ terms of G.P

$Sum=\frac{a(1-r^n)}{1-r}$

Therefore,

$\Rightarrow \frac{\frac{x^2(1-x^{2n})}{1-x^2}+\frac{xy(1-x^n{y}^n)}{1-xy}}{x^2\frac{(1-x^{2n})}{1-x^2}+xy\frac{(1-x^n{y}^n)}{1-xy}}$

$\Rightarrow 1$

Hence, this is the answer.