Compute$\sum_{k=2}^{n+5}\frac {1}{k(k-1)}\binom {n+1}{k-2}2^k$ Compute$\sum_{k=2}^{n+5}\frac {1}{k(k-1)}\binom {n+1}{k-2}2^k$ .
Well, here how I tried and got stuck:


Note that
$$\begin{align*}
\frac1{k(k-1)}\binom{n+1}{k-2}&=\frac{(n+1)!}{k(k-1)(k-2)!\big((n+1)-(k-2)\big)!}\\
&=\frac{(n+1)!}{k!(n+3-k)!}\\
&=\frac1{(n+3)(n+2)}\cdot\frac{(n+3)!}{k!(n+3-k)!}\\
&=\frac1{(n+3)(n+2)}\binom{n+3}k\;,
\end{align*}$$
so
$$\sum_{k=2}^{n+5}\frac1{k(k-1)}\binom{n+1}{k-2}2^k=\frac1{(n+3)(n+2)}\sum_{k=2}^{n+5}\binom{n+3}k2^k\;.$$
Now use the binomial theorem and make a couple of adjustments, a bit like what you did in the previous problem.
