Closed form for $\sum_{k=0}^{n} {n \choose k} \dfrac{2^{k+1}}{k+1}$

Question

What's the closed form for the binomial series $$\sum_{k=0}^{n} {n \choose k} \dfrac{2^{k+1}}{k+1}$$

Answer 1

$$\sum_{k=0}^{n} {n \choose k} \dfrac{2^{k+1}}{k+1} = \sum_{k=0}^{n} {n \choose k} \int_0^2 t^k\text{ d}t$$ Swap the integral and sum (justified because the limits are finite by Fubini's theorem) $$\int_0^2{\color{red}{\sum_{k=0}^{n} {n \choose k} t^k}}\text{ d}t=\int_0^2{\color{red}{(t+1)^n}}\text{ d}t = \frac{3^{n+1}-1}{n+1}$$ Red is just the binomial series.

Answer 2

Without using Calculus,

As $(k+1)\cdot k!=(k+1)!$ for integer $k\ge0,$

$$\dfrac{\binom nk}{k+1}=\dfrac{(n+1)!}{(n+1)\cdot(n+1-(k+1)!)\cdot(k+1)!}=\dfrac{\binom{n+1}{k+1}}{n+1}$$

$$\implies\sum_{k=0}^n\dfrac{a^{k+1}\binom nk}{k+1}=\dfrac1{(n+1)}\cdot\sum_{k=0}^n\binom{n+1}{k+1}a^{k+1}=\dfrac{(a+1)^{n+1}-\binom{n+1}0}{n+1}$$ using $$(x+y)^m=\sum_{r=0}^m\binom mrx^{m-r}y^r$$ for positive integer $m$