Say i got $\displaystyle{\frac{(1-2x)}{(1+3x)^3}}$
I used $\displaystyle{\frac{1}{(1+3x)}}$ $=\sum_{n=0}^\infty(-3)^n x^n$ and differentiated twice
I got $\displaystyle{\frac{(1-2x)}{(1+3x)^3}}$ = $=\sum_{n=0}^\infty [((n)(n-1)(-3)^n)/18] x^{(n-2)}$
Multiply (1-2x) on both side I got
= $\sum_{n=0}^\infty [((n)(n-1)(-3)^n)/18] x^{(n-2)}$ - $2\sum_{n=0}^\infty [((n-1)(n-2)(-3)^{(n-1)}))/18] x^{(n-2)}$
$=\sum_{n=0}^\infty [(5n-4)(n-1)(-3)^n /54 ] x^{(n-2)}$
Is that correct ? I had a feeling that its wrong...
Coefficient of $z^{(n-2)}$ is $\displaystyle{\frac{(5n-4)(n-1)(-3)^n}{54}}$?