Binomial Identity problem I've been working on this binomial identity problem for hours but I seriously have no clue how to deal with this. Here's the problem:
Evaluate the sum
$$1 + 2C(n,1) + \cdots (k + 1)C(n,k) + \cdots + (n + 1)C(n, n)$$
by breaking this sum into two sums, each of which is an identity in this
section.
(I don't even understand what the question actually means - "by breaking this sum into two sums?")
Please please please someone help me. this is killing me inside :(
 A: Another way
$$
\begin{align}
\sum_{k=0}^n (k+1)\binom{n}{k}&=\sum_{k=1}^nk\binom{n}{k}+\sum_{k=0}^n\binom{n}{k}\\
&=\sum_{k=1}^n n\binom{n-1}{k-1}+(1+1)^n\tag{1}\\
&=n\sum_{k=1}^n\binom{n-1}{k-1}+2^n\\
&=n\sum_{u=0}^{n-1}\binom{n-1}{u}
+2^n\\
&=n2^{n-1}+2^n
\end{align}
$$
where in (1) we used the identities
$$
\sum_{k=0}^n\binom{n}{k}=2^n
$$
a consequence of the binomial theorem and
$$
k\binom{n}{k}=n\binom{n-1}{k-1}\quad(n\geq k\geq1)
$$
A: Hint:
$$\sum_{j=0}^n \binom njx^j\equiv(1+x)^n$$
What happens if you differentiate both sides of the above identity wrt $x$ and set $x=1$ ?

Edit:
It seems your sum is $\sum\limits_{k=0}^n(k+1)\binom nk=2^n+\sum_{k=0}^nk\binom nk$ where you can evaluate the latter sum using the above hint.

Here's one algebraic approach similar to the one Gauss used to calculate the sum of the first $n$ natural numbers.
Let $S:=\sum\limits_{j=0}^n j\binom nj$. Then, since $\binom nj=\binom n{n-j}$, we have,
$$2S=\sum_{j=0}^nj\binom nj+\sum_{j=0}^n(n-j)\binom nj=n\sum_{j=0}^n\binom nj=n2^n\implies S=n2^{n-1}$$
