Find the coefficient using binomial theorem. What is the coefficient of $x^{20}$ in the expression: $$(x+1)^{10}.(x^2 -1)^8$$ 

From the binomial theorem you know that
$$(x+1)^{10}=\sum_{k=0}^{10}\binom{10}kx^k$$
and
$$(x^2-1)^8=\sum_{k=0}^8\binom8k(-1)^{8-k}x^{2k}=\sum_{k=0}^8\binom8k(-1)^kx^{2k}\;,$$
so 
$$\begin{align*}
(x+1)^{10}(x^2-1)^8&=\left(\sum_{k=0}^{10}\binom{10}kx^k\right)\left(\sum_{k=0}^8\binom8k(-1)^kx^{2k}\right)\\\\
&=\left(1+10x+\ldots+10x^9+x^{10}\right)\left(1-8x^2+\ldots-8x^{14}+x^{16}\right)\;.
\end{align*}$$
The $x^{20}$ term in the product will be the sum of all terms of the form $(ax^k)(bx^\ell)$ such that $k+\ell=20$. Find all of those combinations of $k$ and $\ell$, and you’re nearly there.
