Closed form for $\sum_{k=0}^{j}\binom{n}{k}^2 x^k$

Question

I need a closed form for the sum $$\sum_{k=0}^{j}\binom{n}{k}^2 x^k$$ where $0<x<1$

Let $y=x^{1/2}$. The above expression can be rewritten as

$$\int_0^{1 } \left(\sum_{k=0}^j \binom{n}{k}y^k e^{2\pi ikt} \right) \left(\sum_{k=0}^j \binom{n}{k}y^k e^{-2\pi ikt} \right) \ dt.$$ Any help would be appreciated. Thank you!

Answer 1

Maple does this sum in terms of hypergeometrics: $$ \sum_{k=0}^{j}\binom{n}{k}^2 x^k = \\ \;{}_2F_1\left(-n,-n;1;x\right) -\binom{n}{j+1}^2 x^{j+1}\;{}_3F_2\left(1,-n+j+1,-n+j+1;2+j,2+j;x\right) $$