<s>[INST] Express in terms of $x$ and $y$ when the values of $x$ and $y$ are given. Given,
$x=1+3a+6a^2+10a^3+\ldots$
$y=1+4b+10b^2+20b^3+\ldots$
$s=1+3ab+5(ab)^2+7(ab)^3+\ldots$
Express $s$ in terms of $x$ and $y$.  
My work:
I could see how the first sequence works, but could not find how the second sequence works, until I wrote down the Pascal's Triangle. Then I realised that,
$x=1+{3 \choose 1}a+{4 \choose 2}a^2+{5 \choose 3}a^3+\ldots$
$y=1+{4 \choose 1}b+{5 \choose 2}b^2+{6 \choose 3}b^3+\ldots$
And $s$ sequence was easy to see. It was just an arithmetico-geometric progression.
$s~~~~~~~=1+3ab+5(ab)^2+7(ab)^3+9(ab)^4+\ldots$
$s(ab)=~\cdot+~~ab+3(ab)^2+5(ab)^3+7(ab)^4+\ldots$
$s(1-ab)=1+2\{ab+(ab)^2+(ab)^3+\ldots\}$
But, I do not see how does that help me. Please help.
 [/INST] Assume, $|a|,|b|<1$
You can find s by using formula for infinite GP in the last step.
$s=\frac{1+\frac{2ab}{1-ab}}{1-ab}=\frac{1+ab}{(1-ab)^2}$
Also note that taylor series for $\frac 1 {(1-x)^n}$ is :
$ 1+{n\choose 1} x +{n+1\choose2}x^2 + {n+2\choose3}x^3... $
Hence $x$ stands for $n=3$ and $y$ for $n=4$
Now find $x$ and $y$ in terms of $a$ and $b$ and put in $s$.
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