Find the coefficient using binomial theorem. What is the coefficient of $x^{20}$ in the expression: $$(x+1)^{10}.(x^2 -1)^8$$ 
 A: $$(x+1)^{10}.(x^2 -1)^8=(x+1)^{18}.(x -1)^8=\left(\sum_{k=0}^{18} {18 \choose k} x^k\right)\cdot\left(\sum_{\ell=0}^8 {8 \choose \ell}x^\ell(-1)^{8-\ell}\right)$$ 
hence the coefficient of $x^{20}$ is
$$\sum_{k+\ell=20}(-1)^{\ell}{18 \choose k}{8 \choose \ell}$$
A: 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.
A: $$(x+1)^{10}.(x^2 -1)^8=\sum_{i=0}^{10}\binom{10}{i}x^i\sum_{j=0}^{8}\binom{8}{j}x^{2j}(-1)^{8-j}$$
$$i+2j=20$$
$$(i,j)=(10,5),(8,6),(6,7),(4,8)$$
$$\binom{10}{10}\binom{8}{5}(-1)^{3}+\binom{10}{8}\binom{8}{6}(-1)^{2}+\binom{10}{6}\binom{8}{7}(-1)+\binom{10}{4}\binom{8}{8}(-1)^0=$$ 
$$=-\binom{8}{5}+\binom{10}{8}\binom{8}{6}-\binom{10}{6}\binom{8}{7}+\binom{10}{4}=-266$$ 
