If $x=1+2a+3a^{2}+4a^{2}+\dots(-1 < a < 1)$ and
$y=1+3b+6b^{2}+10b^{3}+\dots(-1 < b < 1),$
then find $1+ab+(ab)^{2}+(ab)^{3}+\dots$ in terms of $x$ and $y$.
- $\frac{x^{1/2} y^{1/3}}{x^{1/2}+y^{1/3}-1}$
- $\frac{xy}{x+y-1}$
- $\frac{x^{1/3}y^{2/3}}{x^{1/3}+y^{1/2}-1}$
- None of these.