An algebra problem by Rahil Sehgal

Let $u = x^{2019}$ . Then when the power of $x$ is odd, the power of $u$ is also odd. And we have:

$\begin{aligned} \left(1+u\right)^{20} & = {20 \choose 0} + {20 \choose 1}u + {20 \choose 2}u^2 + {20 \choose 3}u^3 + \cdots + {20 \choose 20}u^{20} \\ \left(1-u\right)^{20} & = {20 \choose 0} - {20 \choose 1}u + {20 \choose 2}u^2 - {20 \choose 3}u^3 + \cdots + {20 \choose 20}u^{20} \\ \left(1+u\right)^{20} - \left(1+u\right)^{20} & = 2 \left({20 \choose 1}u + {20 \choose 3}u^3 + {10 \choose 5}u^5 + \cdots + {20 \choose 19}u^{19} \right) \\ \frac {\left(1+u\right)^{20} - \left(1+u\right)^{20}}2 & = {20 \choose 1}u + {20 \choose 3}u^3 + {10 \choose 5}u^5 + \cdots + {20 \choose 19}u^{19} & \small \color{#3D99F6} \text{Putting }u=1 \\ \frac {(2)^{20} - (0)^{20}}2 & = {20 \choose 1} + {20 \choose 3} + {10 \choose 5} + \cdots + {20 \choose 19} \end{aligned}$

Note that the LHS is the sum of coefficients of odd powers of $x$ , which is $= 2^{19} = \boxed{524288}$ .

It will be the sum of odd combinations of 20, i.e.

$\large \displaystyle \sum_{n=0}^9 C_{2n+1}^{20}$

$\large \displaystyle = \frac12 \cdot 2^{20}$

$\color{#3D99F6} = \boxed{\large \displaystyle 524288}$