Binomial sum involving power of $2$ 
Finding $\displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)(k+2)}\cdot 2^{k+2}\binom{n}{k}$

Try: $$\int^{x}_{0}(1+x)^n=\int^{x}_{0}\bigg[\binom{n}{0}+\binom{n}{1}x+\cdots\cdots +\binom{n}{n}x^n\bigg]dx$$
$$\frac{(1+x)^{n+1}-1}{n+1}=\binom{n}{0}x+\binom{n}{1}\frac{x^2}{2}+\cdots \cdots +\binom{n}{n}\frac{x^{n+1}}{n+1}$$
Could some help me to solve it , Thanks

$$f(x)=(1+x)^n =\displaystyle \sum^{n}_{k=0}x^{k}\binom{n}{k} $$
Thus
$$g(x)=\int f(x) dx = \frac{(1+x)^{n+1}-1}{n+1}= \displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)}\cdot x^{k+1}\binom{n}{k}$$
Where the constant of integration was chosen so that $g(0)=0$. Also,
$$\int g(x) dx= \frac{(1+x)^{n+2}-1}{(n+1)(n+2)}-\frac{x}{n+1}=\displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)(k+2)}\cdot x^{k+2}\binom{n}{k}$$
So that $$\displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)(k+2)}\cdot 2^{k+2}\binom{n}{k} =  \frac{(1+2)^{n+2}-1}{(n+1)(n+2)}-\frac{2}{n+1}=  \frac{3^{n+2}-2n-5}{(n+1)(n+2)}$$
