How to find coefficient of $x^8$ in $\frac{1}{(x+3)(x-2)^2}$ Using which way can we find coefficient of $x^8$ in $\frac{1}{(x+3)(x-2)^2}$? I have used binomial theorem but failed to find an answer for it.
 A: $$\frac{1}{(x+3)(x-2)^2}=\frac{1}{3(1+x/3)}\frac{1}{-2(1-x/2)}\frac{1}{-2(x-1/2)}=$$
$$=\frac{1}{12}\sum_{i=0}^{\infty}(-x/3)^i\sum_{j=0}^{\infty}(x/2)^j\sum_{k=0}^{\infty}(x/2)^k=\frac{1}{12}\sum_{l=0}^{\infty}\sum_{i+j+k=l}\frac{x^{l}}{(-3)^i2^{j+k}}$$
from above for $l=8$ we get
$$\frac{1}{12}\sum_{i+j+k=8}\frac{1}{(-3)^i2^{j+k}}$$
A: You can decompose
$$
\frac{1}{(x+3)(x-2)^2}=\frac{A}{1+x/3}+\frac{B}{1-x/2}+\frac{C}{(1-x/2)^2}
$$
and it's just a matter of finding $A$, $B$ and $C$. For $x=0$, $x=1$ and $x=-1$, we get
$$
\begin{cases}
\frac{1}{12}=A+B+C\\
\frac{1}{4}=\frac{3}{4}A+2B+4C\\
\frac{1}{18}=\frac{3}{2}A+\frac{2}{3}B+\frac{4}{9}C
\end{cases}
$$
or
$$
\begin{cases}
A+B+C=1/12\\
3A+8B+16C=1\\
27A+12B+8C=1
\end{cases}
$$
This gives
$$
A=\frac{1}{75},\quad B=\frac{1}{50},\quad C=\frac{1}{20}
$$
(if I computed right).
Now you can recall that
$$
\frac{1}{1+x/3}=\sum_{k\ge0}(-1)^k\frac{x^k}{3^k},
\qquad
\frac{1}{1-x/2}=\sum_{k\ge0}\frac{x^k}{2^k},
$$
and that $1/(1-x/2)^2$ is proportional to the derivative of $1/(1-x/2)$, precisely
$$
\left(\frac{1}{1-x/2}\right)'=\frac{1}{2}\frac{1}{(1-x/2)^2}.
$$
Therefore
$$
\frac{1}{(1-x/2)^2}=2\sum_{k\ge0}(k+1)\frac{x^k}{2^k}
$$
and the coefficient you're looking for is
$$
A\frac{1}{3^8}+B\frac{1}{2^8}+2C\frac{9}{2^8}.
$$
A: use $$(1-x)^{-1}=1+x+x^2 +x^3...$$
and $$(1+x)^{-1}=1-x+x^2-x^3....$$
your question is of the form
$$3^{-1}.(-2)^{-2}(\frac{x}{3}+1)^{-1}(1-\frac{x}{2})^{-2}$$
generally
$$(1-x)^{-k}=1+kx+\frac{k(k+1)}{2!}x^2+ \frac{k(k+1)(k+2)}{3!}x^3+.... $$
now to find coefficient of $x^8$ . group all possible terms so that the resulting degree of the term is $8$.
i think you should get $0.002706025...$
