I know the simplified form of $\sum\limits_{k=0}^{n}{n\choose k}\cdot a^k\cdot b^k\cdot \frac{1}{k+1}$, which is $\frac{(a+b)^{n+1}-b^{n+1}}{a\cdot (n+1)}$, I am wandering if there exists the simplified form of $\sum\limits_{k=0}^{n}{n\choose k}\cdot a^k\cdot b^k\cdot \frac{1}{(k+1)^2}$?
EDIT：
Sorry that it should be $\sum\limits_{k=0}^{n}{n\choose k}\cdot a^k\cdot b^{n-k}\cdot \frac{1}{(k+1)^2}$, my mistake.
On
$$ {\rm f}\left(\mu\right) \equiv \sum_{k = 0}^{n}{n \choose k}{\mu^{k + 1} \over \left(k + 1\right)^{2}}\,, \quad {\rm f}'\left(\mu\right) = \sum_{k = 0}^{n}{n \choose k}{\mu^{k} \over k + 1}\,, \quad \left[\mu{\rm f}\left(\mu\right)\right]' = \sum_{k = 0}^{n}{n \choose k}\mu^{k} = \left(1 + \mu\right)^{n} $$
$$ \mu{\rm f}'\left(\mu\right) = {\left(1 + \mu\right)^{n + 1} - 1 \over n + 1}\,, \qquad {\rm f}\left(\mu\right) = {1 \over n + 1} \int_{0}^{\mu}{\left(1 + \mu'\right)^{n + 1} - 1 \over \mu'}\,{\rm d}\mu' $$
$$ \color{#ff0000}{\sum_{k = 0}^{n}{n \choose k}{a^{k}b^{k} \over \left(k + 1\right)^{2}}} = {1 \over ab} \sum_{k = 0}^{n}{n \choose k} {\left(ab\right)^{k + 1} \over \left(k + 1\right)^{2}} = \color{#ff0000}{{1 \over \left(n + 1\right)ab} \int_{0}^{ab}{\left(1 + \mu\right)^{n + 1} - 1 \over \mu}\,{\rm d}\mu} $$ After integration by parts the integration is reduced to a limit of a Incomplete Beta function derivative. This is related to a hypergeometric one.
No simple form, this summation tends to be a generalized hypergeometric function.