Binomial Identity problem

Question

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 :(

Answer 1

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 $$ k\binom{n}{k}=n\binom{n-1}{k-1}\quad(n\geq k\geq1) $$

Answer 2

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 $x=1$ ?

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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.

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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,

$$2S=\sum_{j=0}^nj\binom nj+\sum_{j=0}^n(n-j)\binom nj=n\sum_{j=0}^n\binom nj=n2^n\implies S=n2^{n-1}$$