Computing $2 binom{n}{0} + 2^2 frac{binom{n}{1}}{2} + 2^3 frac{binom{n}{2}}{3} + cdots + 2^{n+1} frac{binom{n}{n}}{n+1}$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},} newcommand{braces}[1]{leftlbrace,{#1},rightrbrace} newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack} newcommand{dd}{mathrm{d}} newcommand{ds}[1]{displaystyle{#1}} newcommand{expo}[1]{,mathrm{e}^{#1},} newcommand{ic}{mathrm{i}} newcommand{mc}[1]{mathcal{#1}} newcommand{mrm}[1]{mathrm{#1}} newcommand{pars}[1]{left(,{#1},right)} newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}} newcommand{root}[2][]{,sqrt[#1]{,{#2},},} newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}} newcommand{verts}[1]{leftvert,{#1},rightvert}$ begin{align} sum_{k = 0}^{n}{2^{k + 1} over k + 1}{n choose k} & = 2sum_{k = 0}^{n}2^{k}{n choose k}int_{0}^{1}t^{k},dd t = 2int_{0}^{1}sum_{k = 0}^{n}{n choose k}pars{2t}^{k},dd t \[5mm] & = 2int_{0}^{1}pars{1 + 2t}^{n},dd t = left. {pars{1 + 2t}^{n + 1} over n + 1},rightvert_{ 0}^{ 1} = bbx{3^{n + 1} - 1 over n + 1}\ & end{align}

The hint given by @ThomasAndrews indicates a purely algebraic approach.

We obtain begin{align*} color{blue}{sum_{j=0}^nfrac{2^{j+1}}{j+1}binom{n}{j}} &=frac{1}{n+1}sum_{j=0}^n2^{j+1}binom{n+1}{j+1}tag{1}\ &=frac{1}{n+1}sum_{j=1}^{n+1}2^jbinom{n+1}{j}tag{2}\ &,,color{blue}{=frac{1}{n+1}left(3^{n+1}-1right)}tag{3} end{align*}

Comment:

• In (1) we use the binomial identity $frac{n+1}{j+1}binom{n}{j}=binom{n+1}{j+1}$.

• In (2) we shift the index by one and start with $j=1$.

• In (3) we apply the binomial theorem.

The expression looks like it has something to do with binomial expansion of $left(x+1right)^n=sum_{k=0}^{n}{binom{n}{k}x^k}$ evaluated at $x=2$ but with each term being integrated. So we need to integrate both sides with respect to $x$ to get $frac{left(x+1right)^{n+1}}{n+1}+c=sum_{k=0}^{n}{binom{n}{k}frac{x^{k+1}}{k+1}}$
Putting $x=0$ we get $c=frac{-1}{n+1}$.

Finally we need to evaulate the expression at $x=2$ Which will make the sum equal to $frac{left(3right)^{n+1}}{n+1}-frac{1}{n+1}$

Your expression is $displaystylesum_{k=0}^n frac{x^{k+1}}{k+1} binom{n}{k}$ evaluated at $x=2$.

Hint. Can you think of where else $displaystylefrac{x^{k+1}}{k+1}$ shows up (specifically, in calculus)?