Find the coefficient of ${x^9}$ in the expansion of $\left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^3}} \right)..\left( {1 + {x^{100}}} \right)$. The official answer is 8.
How do I find the general term,
Dividing the above equation by $(1-x)$ is not generating the required result.
Hint:
$$\begin{align}9=9+0\\=8+1\\=7+2\\=6+3\\=6+2+1\\=5+4\\=5+3+1\\=4+3+2\end{align}$$
We don't have to worry about 4 summands, since $1+2+3+4>9$. There is no known closed form for the general term.
An "x^9" occurs in this expression exactly when a sequence of powers of $x$s add up to a total exponent of $9$. For instance, the first and eighth terms contribute $x$ and $x^8$, for a total of $x^9$ (where in all other terms you multiply by $1$).
The sequences we care about
So now you can just start writing them down:
9
1,8
2,7
3,6
4,5
1,2,6
1,3,5
2,3,4
... and there are eight of those.
