Don't Expand

Let $u = x^{2017}$ . Then when the power of $x$ is odd, the power of $u$ is also odd. And we have:

$\begin{aligned} \left(1+u\right)^{10} & = {10 \choose 0} + {10 \choose 1}u + {10 \choose 2}u^2 + {10 \choose 3}u^3 + \cdots + {10 \choose 10}u^{10} \\ \left(1-u\right)^{10} & = {10 \choose 0} - {10 \choose 1}u + {10 \choose 2}u^2 - {10 \choose 3}u^3 + \cdots + {10 \choose 10}u^{10} \\ \left(1+u\right)^{10} - \left(1+u\right)^{10} & = 2 \left({10 \choose 1}u + {10 \choose 3}u^3 + {10 \choose 5}u^5 + {10 \choose 7}u^7 + {10 \choose 9}u^9 \right) \\ \frac {\left(1+u\right)^{10} - \left(1+u\right)^{10}}2 & = {10 \choose 1}u + {10 \choose 3}u^3 + {10 \choose 5}u^5 + {10 \choose 7}u^7 + {10 \choose 9}u^9 & \small \color{#3D99F6} \text{Putting }u=1 \\ \frac {(2)^{10} - (0)^{10}}2 & = {10 \choose 1} + {10 \choose 3} + {10 \choose 5} + {10 \choose 7} + {10 \choose 9} \end{aligned}$

Note that the LHS is the sum of coefficients of odd powers of $x$ , which is $= 2^9 = \boxed{512}$ .

Let $f(x) = (x^{2017} + 1)^{10}$

To find the sum of the coefficients of the odd powers of $x$ , just find the value of $\dfrac{f(1) - f(-1)}{2}$ . (proof is left as an exercise to the reader)

here $f(1) = 2^{10}$ and $f(-1) = 0$ giving us the value $2^9 = \boxed {512}$

It will be the sum of odd combinations of 10, i.e.

$\large \displaystyle \sum_{n=0}^4 C_{2n+1}^{10}$

$\large \displaystyle = \frac12 \cdot 2^{10}$

$\color{#3D99F6} = \boxed{\large \displaystyle 512}$