<s>[INST] Sum of Beta functions. Given the summation
$$ S_{n} = \sum_{k=0}^{n-1} \binom{n+k-1}{k} \, ( B(a+k, b+n) + B(a+n, b+k) )$$
is it possible to show that it equals the suspected value of $B(a, b)$?
For the first two $n$ values it can be shown that
\begin{align}
S_{1} &= B(a, b+1) + B(a+1, b) = B(a,b) \\
S_{2} &= B(a, b+2) + B(a+2, b) + 2 \, ( B(a+1, b+2) + B(a+2, b+1) ) \\
&= \frac{\Gamma(a) \, \Gamma(b)}{\Gamma(a+b+2)} \, ((a+b)^2 + (a+b)) = B(a,b)
\end{align}
which leads to the suspected value that $S_{n} = B(a,b)$. Are there other ways that this can be shown to be true? 
 [/INST] A solution using induction. We have $S_n=\int_0^1 x^{a-1}(1-x)^{b-1}Q_n(x)\,dx$, where \begin{align}
Q_n(x)&:=(1-x)^n P_n(x)+x^n P_n(1-x),\\
P_n(x)&:=\sum_{k=0}^{n-1}\binom{n+k-1}{k}x^k,\\
P_{n+1}(x)&=\sum_{k=0}^{n}\binom{n+k-1}{k}x^k+\sum_{k=1}^{n}\binom{n+k-1}{k-1}x^k
\\&=P_n(x)+\binom{2n-1}{n}x^n+x\left(P_{n+1}(x)-\binom{2n}{n}x^n\right),
\end{align}
hence $(1-x)P_{n+1}(x)=P_n(x)+\binom{2n}{n}\left(\frac{x^n}{2}-x^{n+1}\right)$ and $$Q_{n+1}(x)=Q_n(x)+\binom{2n}{n}x^n(1-x)^n\left(\frac12-x+\frac12-(1-x)\right)=Q_n(x).$$
Thus, $Q_n(x)=Q_1(x)\equiv 1$ and $S_n=\int_0^1 x^{a-1}(1-x)^{b-1}\,dx=\mathrm{B}(a,b)$.
</s>