Prove $\sum_{r=0}^n \binom{n}{r} \binom{n+r}{r} (-2)^r =(-1)^n\sum_{r=0}^n \binom{n}{r}^2 2^r$ Here is a combinatorics question that I am struggling.
Prove $\sum_{r=0}^n \binom{n}{r} \binom{n+r}{r} (-2)^r =(-1)^n\sum_{r=0}^n \binom{n}{r}^2 2^r$
I tried to simply the binomial coefficients on both sides but it does not help.
Is there any possible way to prove it?
Thanks for any comments.  
 A: Start with the LHS
$$\sum_{r=0}^n {n\choose r} {n+r\choose r} (-1)^r 2^r
= \sum_{r=0}^n {n\choose r} {n+r\choose n} (-1)^r 2^r
\\ = \sum_{r=0}^n {n\choose r} (-1)^r 2^r
[z^n] (1+z)^{n+r}
= [z^n] (1+z)^n
\sum_{r=0}^n {n\choose r} (-1)^r 2^r (1+z)^r
\\ = [z^n] (1+z)^n (1-2(1+z))^n
= [z^n] (1+z)^n (-1-2z)^n
\\ = (-1)^n [z^n] (1+z)^n (1+2z)^n.$$
We get for the RHS
$$(-1)^n \sum_{r=0}^n {n\choose r}^2 2^r
= (-1)^n \sum_{r=0}^n {n\choose r} {n\choose n-r} 2^r
\\ = (-1)^n \sum_{r=0}^n {n\choose r} 2^r [z^{n-r}] (1+z)^n
= (-1)^n [z^n] \sum_{r=0}^n {n\choose r} 2^r z^r (1+z)^n
\\ = (-1)^n [z^n] (1+z)^n \sum_{r=0}^n {n\choose r} 2^r z^r
= (-1)^n [z^n] (1+z)^n (1+2z)^n.$$
The  two are  identical  and we  may conclude.  (The  second one  also
follows by inspection, we have demonstrated the method here.)
A: There are two representations for the Legendre polynomials; you can find them on the wiki:
$$P_n(x)=\sum_{k=0}^n\binom{n}{k}\binom{n+k}{k}\Big(\frac{x-1}{2}\Big)^k $$
$$P_n(x)=\frac{1}{2^n}\sum_{k=0}^n\binom{n}{k}^2\big(x-1\big)^{n-k} \big(x+1\big)^k\,.$$
Let $x=-3$ in both.
