Would there be any simple equivalent expression for
$n\sum^n_{k=0}{n\choose k}^2a^{n-k}b^k-n\sum^{n-1}_{k=0}{n-1\choose k}^2a^{n-1-k}b^k$?
Both $a$ and $b$ are numbers in $[0,1]$.
I tried to use Legendre polynomial such as $\sum^{n}_{k=0}{n \choose k}^2p^k=(1-p)^nP_n(\frac{1+p}{1-p})$, but I can't come up with one meaningful..
I think that you face hypergeometric functions since $$S_1=\sum^n_{k=0}{n\choose k}^2a^{n-k}\,b^k=a^n \, _2F_1\left(-n,-n;1;\frac{b}{a}\right)$$ $$S_2=\sum^{n-1}_{k=0}{n-1\choose k}^2a^{n-1-k}\,b^k=a^{n-1} \, _2F_1\left(1-n,1-n;1;\frac{b}{a}\right)$$ Then, combine; I do not think that this could further simplify. $$S=n\sum^n_{k=0}{n\choose k}^2a^{n-k}b^k-(n-1)\sum^{n-1}_{k=0}{n-1\choose k}^2a^{n-1-k}b^k=nS_1-(n-1)S_2$$