<s>[INST] 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?
 [/INST] You can write
$$p(x)=(1+10x-9x)(1+10x+9x)\cdots (1+10x-x)(1+10x+x)=((1+10x)^2-(9x)^2)((1+10x)^2-(7x)^2)\cdots ((1+10x)^2-x^2)=(19x^2+20x+1)(51x^2+20x+1)(75x^2+20x+1)(91x^2+20x+1)(99x^2+20x+1)$$
Now you can see that the main coefficients of $x^2$ sum to $19+51+75+91+99=335$ now we can look $\mod x^2$ to get the remaining coefficient of $x^2$ we get  $(20x+1)^5$ and the coefficient of the $x^2$ is $4000$ summing those two you get $4335$.
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