A summation identity I have encountered this identity in Page 616 of Mathematical Methods for Students of Physics and Related Fields (Second Edition) by Sadri Hassani:
$$
\sum_{m = 0}^{n}\left(-1\right)^{m}\,
{\left(\,{2n + 2m}\,\right)!
\over \left(\,{n + m}\,\right)!\,\left(\,{n - m}\,\right)!\,
\left(\,{2m}\,\right)!} = \left(\,{-4}\,\right)^n
$$.
I don't know how one can obtain it directly, however, I tried to prove it by induction.
Thus, for $n = 1$, the identity is valid. If we assume its validity for $n$, we have to show that
$\sum_{m = 0}^{n + 1} (-1)^m \frac{(2n + 2m + 2)!}{(n + m + 1)! (n - m + 1)! (2 m)!} = (-4)^{n + 1}$.
The thing that comes to one's mind is that to separate the ($n + 1$)th term in the left-hand side of the above, and write it as
$(-1)^{n + 1} \frac{(4n + 4)!}{(2n + 2)! (2n + 2)!} + \sum_{m = 0}^{n} (-1)^m \frac{(2n + 2m + 2)!}{(n + m + 1)! (n - m + 1)! (2 m)!}$,
which with a little simplification, it becomes
$(-1)^{n + 1} \frac{(4n + 4)!}{(2n + 2)! (2n + 2)!} + 2 \sum_{m = 0}^{n} (-1)^m \frac{(2n + 2m + 1) (2n + 2m)!}{(n - m + 1) (n + m)! (n - m)! (2 m)!}$.
It seems to me that one cannot simplify it more in order to be able to use the assumption; one could divide $\frac{2n + 2m + 1}{n - m + 1}$ but it doesn't seem to lead anywhere. Any help to proceed from here is appreciated!
 A: Here is a very different kind of answer than what you want. This is a sum of hypergeometric terms, and therefore in principle your identity is algorithmic to verify.
If we let $S(n)$ be your sum, then Zeilberger's algorithm (with, say, SumTools[Hypergeometric][Zeilberger](S, n, m, E) in Maple) returns the result
$$
[E + 4,\ {\frac {m \left( 2\,m-1 \right)  \left( -1 \right) ^{m} \left( 2\,n+2\,m \right) !\,
 \left( 8\,n+6 \right) }{ \left( -n+m-1 \right)  \left( n+m \right) !\, \left( n-m \right) !\,
 \left( 2\,m \right) !\, \left( 2\,{n}^{2}+3\,n+1 \right) }}],
$$
where $E$ is the shift operator $ES(n) = S(n + 1)$. This signifies that $S(n)$ satisfies the recurrence $(E + 4) S(n) = 0$, i.e., $S(n + 1) = -4S(n)$. (The second part is to "verify" the results of the algorithm, if you were so inclined.) Since it is easy to check that $S(0) = 1$, it follows that $S(n) = (-4)^n$ for $n \geq 0$.

Here's a more human approach, though still not "directly" evaluating the sum. I liberally use generating functions - see generatingfunctionology for a nice introduction.
Note that your sum is $$S(n) = \sum_{k = 0}^n (-1)^k {2(n + k) \choose n + k} {n + k \choose n - k} = \sum_{k \leq n} (-1)^k {2(n + k) \choose n + k} {n + k \choose n - k}$$ for $n \geq 0$. If we shift the summation variable back by $n$, then $$S(n) = \sum_{k \leq 2n} (-1)^{k - n} {2k \choose k}{k \choose 2n - k}.$$
There is one obvious simplification to try here: $(-1)^k {2k \choose k} = 4^k {-1/2 \choose k}$, which gives
$$S(n) = (-1)^n \sum_{k \leq 2n} 4^k {-1/2 \choose k}{k \choose 2n - k}.$$
The remaining hard part is the sum. The $2n$ is unimportant now (only $2n$ appears in the sum, never $n$ by itself), so let's define
$$R(n) = \sum_{k \leq n} 4^k {-1/2 \choose k} {k \choose n - k}.$$
If we have $R(n)$, then $S(n) = (-1)^n R(2n)$. [There is probably an easy way to evaluate $R(n)$, but I haven't found it yet.]
Let's stop thinking so hard, define $$R(x) = \sum_{n \geq 0} R(n) x^n,$$ and blindly manipulate some sums:
\begin{align*}
    R(x) &= \sum_{n \geq 0} \sum_{k \leq n} 4^k {-1/2 \choose k} {k \choose n - k} x^n \\
         &= \sum_{k \geq 0} \sum_{n \geq k} 4^k {-1/2 \choose k} {k \choose n - k} x^n \\
         &= \sum_{k \geq 0} 4^k {-1/2 \choose k} \sum_{n \geq k} {k \choose n - k} x^n \\
         &= \sum_{k \geq 0} 4^k {-1/2 \choose k} \sum_{n \geq 0} {k \choose n} x^{n + k} \\
         &= \sum_{k \geq 0} 4^k {-1/2 \choose k} x^k (1 + x)^k \\
         &= \sum_{k \geq 0} [4x(1 + x)]^k {-1/2 \choose k} \\
         &= (1 + 4x(1 + x))^{-1/2} \\
         &= \frac{1}{1 + 2x} \\
         &= \sum_{n \geq 0} (-2)^n x^n.
\end{align*}
Therefore $R(n) = (-2)^n$, which finally gives $S(n) = (-1)^n R(2n) = (-4)^n$.
