How to show that $a_{1}=a_{3}=a_{5}=\cdots=a_{199}=0$ Let $$(1-x+x^2-x^3+\cdots-x^{99}+x^{100})(1+x+x^2+\cdots+x^{100})=a_{0}+a_{1}x+\cdots+a_{200}x^{200}$$
show that
$$a_{1}=a_{3}=a_{5}=\cdots=a_{199}=0$$
I have one methods to solve this problem:
Let$$g(x)=(1-x+x^2-x^3+\cdots-x^{99}+x^{100})(1+x+x^2+\cdots+x^{100})$$
Note $$g(x)=g(-x)$$
so $$a_{1}=a_{3}=\cdots=a_{199}=0$$
there exist other methods?

Let $A (x) = (1-x+x^2-x^3+\cdots-x^{99}+x^{100})$ and $B (x) = 1+x+x^2+\cdots+x^{100}$. Then we have $g(x) = A (x) B (x)$. We have $(x + 1) A (x) = x^{101} + 1$ and $(x - 1) B (x) = x ^ {101} - 1$. Then $$g (x) = \frac {x^ {202} - 1} {x^2 - 1} = x^{200} + x^{198} + \cdots + 1,$$ as desired.
