I have the following expression,
$$\sum_{k=0}^{n}\binom{n}{k}\frac{1}{n+1-k}x^ky^{n-k}$$
This seems awfully close to the output of the binomial theorem (for $(x+y)^{n}$). Thus my question is, is the above a known identity/expression? I'd like to get it in a nicer form if possible.
Take your basic binomial theorem $$(x+t)^n=\sum_{k=0}^n \binom nk x^kt^{n-k}\ .$$ Integrate for $t$ from 0 to $y$: $$\frac1{n+1}\bigl((x+y)^{n+1}-x^{n+1}\bigr)=\sum_{k=0}^n \binom nk x^k\frac{y^{n+1-k}}{n+1-k}\ .$$ Divide by $y$: $$\frac1{n+1}\frac{(x+y)^{n+1}-x^{n+1}}y=\langle\hbox{your sum}\rangle\ .$$
Change the variable of summation to $j:=n-k$ and the sum can be written $$ x^n\sum_{j=0}^n{n\choose j}\frac1{j+1}\left(\frac yx\right)^j.\tag1 $$ You can now evaluate the sum in (1) using the identity $$ \sum_{k=0}^n\frac{1}{k+1}{n \choose k}x^k=\frac{(x+1)^{n+1}-1}{(n+1)x} $$ to obtain $$ \sum_{k=0}^{n}\binom{n}{k}\frac{1}{n+1-k}x^ky^{n-k}=\frac{(x+y)^{n+1}-x^{n+1}}{(n+1)y}. $$
