Would there be any simple expression for
$\sum^{n+1}_{k=0}{n+1 \choose k}^2p^k-\sum^{n}_{k=0}{n \choose k}^2p^k$ ?
I tried to use $\sum^{n}_{k=0}{n \choose k}^2p^k=(1-p)^nP_n(\frac{1+p}{1-p})$ where $P_n$ is Legendre polynomial, but I can't find any meaningful expression. I think there's some property of Legendre polynomial I'm missing right now.
$$S_n=\sum^{n+1}_{k=0}{n+1 \choose k}^2p^k-\sum^{n}_{k=0}{n \choose k}^2p^k=\, _2F_1(-n-1,-n-1;1;p)-(1-p)^n P_n\left(\frac{1+p}{1-p}\right)$$ where appears the hypergeometric function. I do not think that this could simplify further.
The only thing which is nice is that $$\lim_{p\to 1} \, S_n=\frac{4^n (3 n+1) \,\Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }\, \Gamma (n+2)}$$
Edit
For very small values of $p$, expansion leads to $$S_n=(2 n+1) p+n^3 p^2+\frac{ (n-1)^2 n^2 (2 n-1)}{12} p^3+\frac{(n-2)^2 (n-1)^3 n^2}{72} p^4+O\left(p^5\right)$$ which seems to work quite well as shown below (using $p=10^{-k}$) $$\left( \begin{array}{cccc} n & k & \text{exact} & \text{approximation} \\ 10 & 2 & 0.323488 & 0.323473 \\ 10 & 3 & 0.0220129 & 0.0220129 \\ 10 & 4 & 0.00211001 & 0.00211001 \\ 10 & 5 & 0.0002101 & 0.0002101 \\ 10 & 6 & 0.000021001 & 0.000021001\\ & & & \\ 20 & 2 & 1.82165 & 1.80276 \\ 20 & 3 & 0.0494818 & 0.0494816 \\ 20 & 4 & 0.00418047 & 0.00418047 \\ 20 & 5 & 0.0004108 & 0.0004108 \\ 20 & 6 & 0.000041008 & 0.000041008 \\ & & & \\ 30 & 2 & 10.4980 & 9.42155 \\ 30 & 3 & 0.0919692 & 0.0919604 \\ 30 & 4 & 0.00637375 & 0.00637375 \\ 30 & 5 & 0.000612704 & 0.000612704 \\ 30 & 6 & 0.000061027 & 0.000061027 \end{array} \right)$$