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:


It is easy to calculate the value of $3\sum\frac{1}{r+1}\binom{n}{r}$ by integrating

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?




Now $\displaystyle\dfrac1{r+1}\binom nr=\dfrac1{n+1}\binom{n+1}{r+1}$

Finally $\displaystyle\sum_{r=0}^m\binom mr=(1+1)^m$

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  • $\begingroup$ This is pretty simple as compared to my method, Thank you! $\endgroup$ – oshhh Feb 21 '17 at 11:19

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