(*) Find the coefficient of x^r , x^4 in the expansion of :
\(\displaystyle \frac{9x^2}{(1+3x)^3}\)
My work so far is :
\(\displaystyle \frac{1}{1+x} \ = \ 1 \ -x \ +x^2 +.... \ (-1)^r*x^r + \ x^{r+2}\)
By differentiating :
\(\displaystyle \frac{-1}{(1+x)^2} \ = \ 0 \ -1 \ ...\ +(-1)^r*r*x^{r-1} \ + \ (r+2)*x^{r+1}\)
\(\displaystyle \frac{2}{(1+x)^3} \ = \ 2 \ ..\ (-1)^r*r*(r-1)*r^{r-2} \ + \ (r+2)(r+1)*x^r.\)
So to have the form of the original , I must multiply by 0.5 then multiply by 9x^2 , then replace x by 3x.
What I got is the following :
\(\displaystyle (9)(1/2)(-1)^r*r*(r-1)*(3^r).\)
Any help will be greatly appreciated.