Problem source

By considering the expansions in powers of $x$
of both sides of the identity
$$ 
{(1+x)^n}{(1+x)^n}\equiv{(1+x)^{2n}},
$$
show that
$$ 
\sum_{s=0}^n {n\choose s}^2 = {2n\choose n},
$$
where $\displaystyle  {n\choose s}= \frac{n!}{s!\,(n-s)!}$.
By considering similar identities, or otherwise, show also that: 
\begin{questionparts}
\item if $n$ is an even integer, then
$\displaystyle \sum_{s=0}^n {{(-1)}^s}{n \choose s}^2= (-1)^{n/2}{n \choose n/2};$
\item 
$\displaystyle \sum\limits_{t=1}^ n 2t { n \choose t}^2 = n {2n\choose n} .$ 
\end{questionparts}

Solution source

To obtain the coefficient of $x^n$ on the RHS we clearly have $\displaystyle \binom{2n}n$. 

To obtain the coefficient of $x^n$ on the LHS we can obtain $x^s$ from the first bracket and $x^{n-s}$ from the second bracket, ie $\displaystyle \sum_{s=0}^n \binom{n}{s}\binom{n}{n-s} = \sum_{s=0}^n \binom{n}{s}\binom{n}{s} = \sum_{s=0}^n \binom{n}{s}^2$

\begin{questionparts}
\item Consider $(1-x)^n(1+x)^n = (1-x^2)^n$, then the coefficient of $x^n$ (if $n$ is even) is for the RHS $\displaystyle (-1)^{n/2} \binom{n}{n/2}$.

For the LHS, we can obtain $x^n$ via $x^s$ and $x^{n-s}$ which is $\displaystyle \sum_{s=0}^n (-1)^s\binom{n}{s}\binom{n}{n-s} = \sum_{s=0}^n (-1)^s\binom{n}{s}^2$

\item Notice that 
\begin{align*}
&& \sum_{t=1}^ n 2t { n \choose t}^2 &= n {2n\choose n} \\
\Leftrightarrow && \sum_{t=1}^ n 2t \frac{n}{t} { n-1 \choose t-1}\binom{n}{t} &= n \frac{2n}{n}{2n-1\choose n-1} \\
\Leftrightarrow && \sum_{t=1}^ n { n-1 \choose t-1}\binom{n}{t} &= {2n-1\choose n-1} \\
\Leftrightarrow && \sum_{t=1}^ n { n-1 \choose t-1}\binom{n}{n-t} &= {2n-1\choose n-1} \\
\end{align*}

but this is exactly what we would obtain by considering the coefficient of $x^{n-1}$ in $(1+x)^{n-1}(1+x)^n \equiv (1+x)^{2n-1}$
\end{questionparts}