If $\sum^n_{k=1} (\frac{k+2}{k+1})^nC_k = \frac{2^n-1}{6}$ then find n. Problem : 
If $\sum^n_{k=1} (\frac{k+2}{k+1})^nC_k = \frac{2^n-1}{6}$ then find n. 
My approach : 
$\sum^n_{k=1} (\frac{k+2}{k+1})^nC_k = \frac{3}{2}^nC_1  +\frac{4}{3}^nC_2 +\frac{5}{4}^nC_3 +\cdots \frac{n+2}{n+1}$ 
Now how to proceed further  please suggest will be of great help thanks. 

We have
$$\sum_{k=1}^n \left(\dfrac{k+2}{k+1}\right) \dbinom{n}k = \sum_{k=1}^n \dbinom{n}k + \sum_{k=1}^n \dfrac1{k+1} \dbinom{n}k$$
We have
$$\sum_{k=0}^n \dbinom{n}k x^k = (1+x)^n \,\,\,\,\, (\spadesuit)$$
Taking $x=1$, we obtain
$$\sum_{k=1}^n \dbinom{n}k x^k = 2^n-1$$
Integrating $(\spadesuit)$, we obtain
$$\sum_{k=0}^n \dfrac1{k+1} \dbinom{n}k x^{k+1} = \dfrac{(1+x)^{n+1}-1}{n+1} \,\,\,\,\, (\clubsuit)$$
Plugging in $x=1$, we obtain
$$\sum_{k=1}^n \dfrac1{k+1} \dbinom{n}k = \dfrac{2^{n+1}-n-2}{n+1}$$
I trust you can finish it off from here.
