Compute$\sum_{k=2}^{n+5}\frac {1}{k(k-1)}\binom {n+1}{k-2}2^k$

Question

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:

Answer 1

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.

Answer 2

\begin{align} \sum_{k=2}^{n+5}\frac {1}{k(k-1)}\binom {n+1}{k-2}2^k&=\sum_{k=0}^{n+3}\frac {1}{(k+2)(k+1)}\binom {n+1}{k}2^{k+2}\\ &=\sum_{k=0}^{n+1}\frac {1}{(k+2)(k+1)}\binom {n+1}{k}2^{k+2}\\ f(x)&=\sum_{k=0}^{n+1}\frac {1}{(k+2)(k+1)}\binom {n+1}{k}x^{k+2}\\ f'(x)&=\sum_{k=0}^{n+1}\frac {1}{k+1}\binom {n+1}{k}x^{k+1}\\ f''(x)&=\sum_{k=0}^{n+1}\binom {n+1}{k}x^{k}=(1+x)^{n+1}\\ f'(x)&=\frac1{n+2}(1+x)^{n+2}+C_0\\ f'(0)&=0 \rightarrow C_0=-\frac1{n+2}2^{n+2}\\ f(x)&=\frac1{(n+2)(n+3)}(1+x)^{n+3}-\frac1{n+2}2^{n+2}x+C_1\\ f(0)&=0\rightarrow C_1=-\frac1{(n+2)(n+3)}2^{n+3}\\ \sum_{k=0}^{n+1}\frac {1}{(k+2)(k+1)}\binom {n+1}{k}2^{k+2}&=f(2)\\ &=\frac1{(n+2)(n+3)}(1+x)^{n+3}-\frac1{n+2}2^{n+2}x-\frac1{(n+2)(n+3)}2^{n+3}\\ &=\frac1{(n+2)(n+3)}3^{n+3}-\frac1{n+2}2^{n+3}-\frac1{(n+2)(n+3)}2^{n+3}\\ \end{align}

Answer 3

with the work above i have got $${\frac {27\,{3}^{n}-2\,n-7}{ \left( n+3 \right) \left( n+2 \right) }}$$