Finding the value of $\sum\frac{2r+3}{r+1}\binom{n}{r}$

If $\sum\frac{2r+3}{r+1}\binom{n}{r}=\frac{(n+k)2^{n+1}-1}{n+1}$ , find the value of $k$.

I am facing problem in simplifying the summation. The way I can think of to approach this question is breaking the summation as:

$$\sum\frac{2r}{r+1}\binom{n}{r}+3\sum\frac{1}{r+1}\binom{n}{r}.$$

It is easy to calculate the value of $3\sum\frac{1}{r+1}\binom{n}{r}$ by integrating
$x(1+x)^n=\binom{n}{0}x+\binom{n}{1}x^2+\binom{n}{2}x^3+\cdots$

But how do I find the value of $\sum\frac{2r}{r+1}\binom{n}{r}$? Can't think of any way and I would require a hint for that.
Is there another simpler way to solve this summation using binomial theorem?

$$\dfrac{2r+1}{r+1}=2-\dfrac1{r+1}$$
Now $\displaystyle\dfrac1{r+1}\binom nr=\dfrac1{n+1}\binom{n+1}{r+1}$
Finally $\displaystyle\sum_{r=0}^m\binom mr=(1+1)^m$