Identities for the product of $\sum_{i=0}^{n} x^i$ and $\sum_{i=0}^{n} (-1)^i x^i$ I saw in a response to a post on this web site that
\begin{equation*}
(1 - x + x^2)(1 + x + x^2) = 1 + x^2 + x^4 .
\end{equation*}
It seems to me that
\begin{equation*}
\sum_{i=0}^{2n} x^{i} \sum_{i=0}^{2n} (-1)^{i} x^{i}
= \sum_{i=0}^{2n} x^{2i} .
\end{equation*}
I understand that this looks like the product of certain partial sums for $1/(1+x)$ and $1/(1-x)$. This does not seem relevant, though. There is another identity of certain partial sums of these functions, and it looks different.
\begin{equation*}
\sum_{i=0}^{2n-1} x^{i} \sum_{i=0}^{2n-1} (-1)^{i} x^{i}
= \sum_{i=0}^{n-1} x^{2i} - \sum_{i=n}^{2n-1} x^{2i} .
\end{equation*}
Without using induction, may someone offer an explanation to these identities?  
 A: Hint:
$$
\frac{x^{2n+1}+1}{x+1} \cdot \frac{x^{2n+1}-1}{x-1} = \frac{(x^2)^{2n+1}-1}{x^2-1}
$$

[ EDIT ]   Hint (for the second identity): using the first and already proved identity, write it as:

$$
\require{cancel}
\begin{align}
\sum_{i=0}^{2n-1} x^{i} \sum_{i=0}^{2n-1} (-1)^{i} x^{i} & = \left(\left(\sum_{i=0}^{2n} x^{i}\right) - x^{2n}\right) \left(\left(\sum_{i=0}^{2n} (-1)^{i} x^{i}\right)-x^{2n}\right) \\[5px]
& = \left(\sum_{i=0}^{2n} x^{i}\right)\left(\sum_{i=0}^{2n} (-1)^{i} x^{i}\right) -x^{2n}\left(\sum_{i=0}^{2n} x^{i} + \sum_{i=0}^{2n} (-1)^{i} x^{i}\right) + x^{4n} \\[5px]
 & = \sum_{i=0}^{2n} x^{2i} - 2 \,x^{2n}\, \sum_{i=0}^{n} x^{2i} + x^{4n} \\[5px]
 & = \left(\sum_{i=0}^{2n-1} x^{2i} + \bcancel{x^{4n}}\right) - 2\,\left(\sum_{i=n}^{2n-1} x^{2i} + \bcancel{x^{4n}}\right) + \bcancel{x^{4n}} = \cdots
\end{align}
$$
A: For an alternative and perhaps more generalizable approach, consider using the Cauchy product formula here
$$\sum_{i=0}^\infty x^i \sum_{j=0}^\infty (-1)^jx^j = \sum_{k=0}^\infty \sum_{n=0}^k (-1)^nx^nx^{k-n} = \sum_{k=0}^\infty x^k \sum_{n=0}^k (-1)^n$$
Breaking this into even and odd parity's of $k$,
$$\sum_{k=0}^\infty x^k \sum_{n=0}^k (-1)^n = \sum_{p=0}^\infty x^{2p} \sum_{n=0}^{2p} (-1)^n + \sum_{p=0}^\infty x^{2p+1} \sum_{n=0}^{2p+1} (-1)^n$$
$$\sum_{n=0}^{2p+1} (-1)^n = 0 \space \text{ and } \space \sum_{n=0}^{2p} (-1)^n = 1$$
as $\sum_{n=0}^{k} (-1)^n = (1 - 1 + 1 - 1 ...) \space\space\space\space k+1$ times. So, if $k$ is odd, there are $\lceil\frac{k}{2} \rceil$ $\space 1's$ and $-1's$, so that the resulting sum is $0$. If $k$ is even, there are $\frac{k}{2} \space\space -1's$ and $(\frac{k}{2} + 1) \space\space 1's$, so the resulting sum is $1$. Thus, 
$$\sum_{k=0}^\infty x^k \sum_{n=0}^k (-1)^n = \sum_{p = 0}^\infty x^{2p}$$
in other words,
$$(1+x+x^2+...)(1-x+x^2-...) = 1 + x^2 + x^4 +...$$
Note that this can be well generalized to things like $x^{2k - 7}$ and $x^{-k}$ as well.
