Help with a summation: $\sum_{r=0}^{n-4}(r+2)\binom{n-4}{r}=n2^{n-5}$ I have been trying to figure out why the summation of $(r+2)\binom{n-4}{r}$ from $r=0$ to $(n-4)$ is equal to $n2^{n-5}$, but I can't seem to get it to work. The only thing I can think of that would be relevant is that the summation of $\binom{n}{k}$ from $k=0$ to $n$ is $2^n$, but I can't get it to follow. Could anyone help me with this, either with a solution or by telling me what method I'm meant to use? Are there by any chance any facts I'm meant to know?
 A: You can decompose your sum as $$\sum_{r=0}^{n-4} r {n-4 \choose r}+\sum_{r=0}^{n-4} 2{n-4 \choose r}$$
The second sum is $2\cdot 2^{n-4}=2^{n-3}$ by the formula you gave.
You also require the result that $$\sum_{r=0}^{n}r{n \choose r}=n2^{n-1}$$
Which gives the first sum as $(n-4)\cdot 2^{n-5}$ 
Your full sum is therefore $2^{n-5}(n-4+4)=n2^{n-5}$   
I don't know if you have proved these identities yourself, but if not, consider the binomial expansions of $(1+x)^n$ and its derivative at $x=1$.
A: Hint: Use the binomial identity
$$r \binom{n-4}{r} = (n-4) \binom{n-5}{r-1}$$
Keep in mind that $\binom{n-5}{-1} = \binom{n-5}{n-4} = 0.$
A: $$\sum_{r=o}^{n-4}(r+2)\binom{n-4}{r}=\sum_{r=0}^{n-4}r\binom{n-4}{r}+2\sum_{r=0}^{n-4}\binom{n-4}{r}=$$
$$=\sum_{r=0}^{n-4}(n-4)\binom{n-5}{r-1}+2\cdot2^{n-4}=(n-4)\sum_{j=0}^{n-5}\binom{n-5}{j}+2^{n-3}=$$
$$=(n-4)2^{n-5}+2^{n-3}=n\cdot 2^{n-5}-4\cdot2^{n-5}+2^{n-3}=$$
$$=n\cdot 2^{n-5}-2^2\cdot2^{n-5}+2^{n-3}=n\cdot 2^{n-5}-2^{n-3}+2^{n-3}=n\cdot 2^{n-5}$$
Using
$$k\binom{n}{k}=n\binom{n-1}{k-1}\Rightarrow r\binom{n-4}{r}=(n-4)\binom{n-5}{r-1}$$
A: First of all,
$$\sum_{r=0}^{n-4}{n-4 \choose r}(r+2)$$
is the same as (where $m = n-4$)
$$\sum_{r=0}^{m}{m \choose r}(r+2) = \sum_{r=0}^{m}{m \choose r}r + 2\sum_{r=0}^{m}{m \choose r}$$
As you've said, using the binomial theorem,
$$2\sum_{r=0}^{m}{m \choose r}1^r 1^{m-r} = 2(1+1)^m = 2^{m+1}$$
The second part, can be explained as having a bunch of people ($m$), in how many ways can you pick a group and designate one of them as a leader:


*

*You either pick a group of size $r$ (${m \choose r}$ ways) and pick one as the leader ($r$ ways), for every group size - Leading to sum:


$$\sum_{r=0}^{m}{m \choose r}r$$


*

*Pick one as the leader ($m$ choices) and for every one of the remainning people ($m-1$) decide whether they are in the group or not ($2^{m-1}$ ways) leading to


$$m 2^{m-1}$$
So that together you have
$$m 2^{m-1} + 2^{m+1} = (m+4) 2^{m-1} = n 2 ^{-5}$$
A: The summation you quote will be useful.
Then note the following trick, which has a vast scope (http://www.math.upenn.edu/~wilf/DownldGF.html).
Start with
\begin{equation}
(1 + x)^{n-4} = \sum_{r=0}^{n-4} \dbinom{n-4}{r} x^r.
\end{equation}
Derive both sides and then set $x = 1$ to get... then use the other summation.
