Sum of sequence of $\binom{n}{r}$ How can we find the sum of $ \binom{21}{1}+ 5\binom{21}{5}+ 9\binom{21}{9}....+17\binom{21}{17}+ 21\binom{21}{21}$?
I have no clue how to begin. I guess complex numbers might help. 
EDIT: Actually the real question was that the sum above was k and we had to find its prime factors. And the answer according to the key involves complex numbers. They have directly written that:
$k=\frac{21( 2^{20} + 0^{20} + (1+i)^{20} + (1-i)^{20})}{4} = 21(2^{18}-2^9)$
I didn't get the above.

Your sum is $$S=\sum_{k=0}^m(4k+1)\binom{n}{1+4k}$$ with $n=21, m=5$. To find this sum we proceed as below $$(1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k\\ \implies n(1+x)^{n-1}=\sum_{k=0}^n k\binom{n}{k}x^{k-1}$$ Let $\alpha=e^{j2\pi/r}$ be the $r$th root of unity. Then,  $$n(1+\alpha^s)^{n-1}=\sum_{k=0}^n k\binom{n}{k}\alpha^{s(k-1)},\quad\ s=0,1,\cdots,\ r-1$$ Adding these $r$ equations we will get $$n\sum_{s=0}^{r-1}(1+\alpha^s)^{n-1}=r\sum_{0\le k\le n: r|k}k\binom{n}{k}\\ \implies =\sum_{0\le k\le n: r|k}k\binom{n}{k}=\frac{n}{r}\sum_{s=0}^{r-1}2^{n-1}e^{j(n-1)\pi s/r}\cos^{n-1} \left(\frac{\pi s}{r}\right)$$ I think you can proceed from here...
