How to compute the summation of a binomial coefficient/ show the following is true $\sum\limits_{k=0}^n \left(2k+1\right) \dbinom{n}{k} = 2^n\left(n+1\right)$.
I know that you have to use the binomial coefficient, but I'm not sure how to manipulate the original summation to make the binomial coefficient useful.

Consider
$$f(x)=(1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k$$
so
$$f'(x)=\sum_{k=1}^{n}k\binom{n}{k}x^{k-1}$$
Hence
$$f'(1)=\sum_{k=1}^{n}k\binom{n}{k}=\sum_{k=0}^{n}k\binom{n}{k}$$
On the other hand
$$f'(x)=n(1+x)^{n-1}\Rightarrow f'(1)=n2^{n-1}$$
It follows that
$$\sum_{k=0}^{n}k\binom{n}{k}=n2^{n-1}$$
Similarly we have
$$\sum_{k=0}^n\binom{n}{k}=f(1)=2^n$$
Put all stuff together gives
$$\sum_{k=0}^n(2k+1)\binom{n}{k}=2^n(n+1)$$
