$\displaystyle{\frac{18}{(1+3x)^3}}$ $=\sum_{n=0}^\infty n(n-1)(-3)^n x^{n-2}$
If i got up to this, how could i get $\displaystyle{\frac{(1-2x)}{(1+3x)^3}}$ ?
When i tried to multiply both side, some people says n-m some says n+m for $x^m$
Could someone kindly show me the working out please?
My working out:
$\displaystyle{\frac{(1-2x)}{(1+3x)^3}}$ = $=\sum_{n=0}^\infty [((n)(n-1)(-3)^n)/18] x^{(n-2)}$
Multiply (1-2x) on both side I got
= $\sum_{n=0}^\infty [((n)(n-1)(-3)^n)/18] x^{(n-2)}$ - $2\sum_{n=0}^\infty [((n-1)(n-2)(-3)^{(n-1)}))/18] x^{(n-2)}$
$=\sum_{n=0}^\infty [(5n-4)(n-1)(-3)^n /54 ] x^{(n-2)}$