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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?

3 Answers3

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HInt 1: $$g(x)=(1+x^2+\cdots+x^{100})^2-x^2(1+x^2+x^4+\cdots+x^{98})^2$$

Hint 2: since $$g(x)=\dfrac{1+x^{101}}{1+x}\cdot\dfrac{1-x^{101}}{1-x}=\dfrac{1-x^{202}}{1-x^2}=1+x^2+\cdots+x^{198}+x^{200}$$

math110
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HINT multiply few terms of first expression with $x^2$ and $x^3$ and observe what gets cancelled out then you can prove it for all even-odd pairs but note even with this 1 and $a_{200}$ only remain till last expression . so you have proved it.I hope you know odd terms of any even degree polynomial are $0$.

Archis Welankar
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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.