Is there a close form for $g(a,b,n)=\sum\limits_{k=0}^{n}\binom{n}{k}\frac{1}{ak+b}$? We can be sure, that for $a>0$, $b>0$
$$f(a,b,n)=\sum\limits_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{ak+b}=\frac{(an)!^{(a)}}{(an+b)!^{(a)}}$$
where $(an+b)!^{(a)}$ denotes multifactorial: $(n)!^{(1)}=n!$, $(2n)!^{(2)}=(2n)!!$, etc.
But if we make a little change
$$g(a,b,n)=\sum\limits_{k=0}^{n}\binom{n}{k}\frac{1}{ak+b}$$
we have
$$\sum\limits_{k=0}^{n}\binom{n}{k}\frac{1}{k+1}=\frac{2^{n+1}-1}{n+1}$$
$$\sum\limits_{k=0}^{n}\binom{n}{k}\frac{1}{k+2}=\frac{n2^{n+1}+1}{(n+1)(n+2)}$$
in general
$$\sum\limits_{k=0}^{n}\binom{n}{k}\frac{1}{k+2c}=\frac{n(2c-1)!}{n^{\overline {2c+1}}}\left(1+2^{n+1}\sum\limits_{m=1}^{c}\frac{n^{\overline {2m-1}}}{(2m-1)!}\right)$$
$$\sum\limits_{k=0}^{n}\binom{n}{k}\frac{1}{k+2c-1}=\frac{n(2c-2)!}{n^{\overline {2c}}}\left(-1+2^{n+1}\sum\limits_{m=1}^{c}\frac{n^{\overline {2m-2}}}{(2m-2)!}\right)$$
Is there a close form for $g(a,b,n)$?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\Large\mrm{f}\pars{a,b,n} = {\large ?}.}$

\begin{align}
\mrm{f}\pars{a,b,n} & \equiv
\sum_{k = 0}^{n}{n \choose k}{\pars{-1}^{k} \over ak + b} =
{1 \over a}\sum_{k = 0}^{n}{n \choose k}\pars{-1}^{k}
\int_{0}^{1}t^{k + b/a - 1}\,\dd t
\\[5mm] & =
{1 \over a}\int_{0}^{1}t^{b/a - 1}
\sum_{k = 0}^{n}{n \choose k}\pars{-t}^{k}\,\dd t =
{1 \over a}\int_{0}^{1}t^{b/a - 1}\pars{1 - t}^{n}\,\dd t
\\[5mm] & =
\bbx{{1 \over a}\,{\Gamma\pars{b/a}\Gamma\pars{n + 1} \over
\Gamma\pars{b/a + n + 1}}\,,\qquad\Re\pars{b \over a} > 0}
\end{align}


$\ds{\Large\mrm{g}\pars{a,b,n} = {\large ?}.}$

In following the $\textsf{'above procedure'}$, I'll arrive to
\begin{align}
\mrm{g}\pars{a,b,n} & \equiv
\sum_{k = 0}^{n}{n \choose k}{1 \over ak + b} =
{1 \over a}\int_{0}^{1}t^{b/a - 1}\pars{1 + t}^{n}\,\dd t
\,\,\,\stackrel{t\ \mapsto\ 1- t}{=}\,\,\,
{1 \over a}\int_{0}^{1}\pars{1 - t}^{b/a - 1}\,\pars{2 - t}^{n}\,\dd t
\\[5mm] & =
{2^{n} \over a}\int_{0}^{1}t^{1 - 1}\pars{1 - t}^{b/a - 1}
\,\pars{1 - {1 \over 2}\,t}^{n}\,\dd t
\end{align}

which is related to the
  Euler Type Expression for the Hypergeometric Function $\ds{\mbox{}_{2}\mrm{F}_{1}}$.

Namely,
\begin{align}
\mrm{g}\pars{a,b,n} & \equiv
\sum_{k = 0}^{n}{n \choose k}{1 \over ak + b} =
{2^{n} \over a}\,\
\overbrace{\mrm{B}\pars{1,{b \over a}}}^{\ds{a \over b}}\
\mbox{}_{2}\mrm{F}_{1}\pars{-n,1;{b \over a} + 1;{1 \over 2}}
\\[5mm] & =
\bbx{{2^{n} \over b}\,
\mbox{}_{2}\mrm{F}_{1}\pars{-n,1;{b \over a} + 1;{1 \over 2}}\,,\qquad
\Re\pars{b \over a} > 0}
\end{align}

$\ds{\mrm{B}}$  is the
  Beta Function.

