Coefficient of $x^2$ in a polynomial 
Let $p(x)$ be an polynomial given by
  $$p(x)=(1+x)(1+3x)(1+5x)\cdots(1+19x)$$ Find the coefficient of $x^2$.

I see that this coefficient is the sum of products of roots of $p(x)$ get eight to eight. But, there is ${10 \choose 8}=45$ terms in this sum. Someone know other way more fast?

Coefficient of $x^2$ in $(1+x)(1+3x)(1+5x)\cdots (1+(2n-1)x)$ is:
$$\begin{align}
&&1\cdot (3+5+7+9+\cdots +(2n-1))&\\
&&+3\cdot (5+7+9+\cdots +(2n-1))&\\
&&+5\cdot (7+9+\cdots +(2n-1))&\\
&&+7\cdot (9+\cdots +(2n-1))&\\
&&\vdots&\\
&&\cdots +(2n-3)\cdot((2n-1))\\
& && =\sum_{i=1}^{n-1}\sum_{j=i+1}^n(2i-1)(2j-1)\\
& && =\sum_{j=2}^n(2j-1)\;\sum_{i=1}^{j-1}(2i-1)\\
& && =\sum_{j=2}^n(2j-1)\;(j-1)^2\\
& && =\sum_{j=1}^{n-1}(2j+1)\;j^2\\
& && =\sum_{j=1}^{n-1}(2j^3+j^2)\\
& && =2\cdot \left[\frac {n(n-1)}2\right]^2+\frac 16(n-1)n(2n-1)\\
& && =\frac 16n(n-1)(3n^2-n-1)
\end{align}$$
Putting $n=10$ gives
$$[x^2]\; (1+x)(1+3x)(1+5x)\cdots (1+19x)=\color{red}{4335}$$
