summation - Simplifying $sumlimits_{k=0}^{infty}frac 1{2^{k+1}}sumlimits_{n=0}^{infty}binom knfrac 1{2(n+1)(3n+1)}$

Question: Is there a way to simplify $$\sum\limits_{k=0}^{\infty}\dfrac 1{2^{k+1}}\sum\limits_{n=0}^{\infty}\binom kn\dfrac 1{2(n+1)(3n+1)}\tag{1}$$

Into a single summation symbol? $\sum\limits_{k=0}^{\infty}\text{something}$

I inputed it into WolframAlpha and got a really complicated expression $$\sum\limits_{k=0}^{\infty}\dfrac {1-2^{k+1}+3\left(_2F_1\left[\begin{array}{c c}\frac 13,-k\\\frac 43\end{array};-1\right]\right)+3k\left(_2F_1\left[\begin{array}{c c}\frac 13,-k\\\frac 43\end{array};-1\right]\right)}{2^{k+3}(k+1)}$$
Which isn't what I really wanted because the inner sum is significantly more complex than before. Is there a way?

I'm still relatively new to this. If you have a hint, it would mean a lot if you commented it!