Two combinatorial identities: $\sum\limits_{k=0}^{n} \binom{n}{k}k2^{n-k}$ and $\sum\limits_{k=1}^{n-2} \binom{n}{k}\frac{k}{n-k}$ I have to compute the following quantity:
$$
1)    \sum\limits_{k=0}^{n} \binom{n}{k}k2^{n-k}
$$
Moreover, I have to give an upper bound for the following quantity:
$$
2)    \sum\limits_{k=1}^{n-2} \binom{n}{k}\frac{k}{n-k}
$$
As regards 1), I see that  $\binom{n}{k}k2^{n-k}=\frac{n! (n-k+1)2^{n-k}}{(n-k+1)!(k-1)!}$, i.e. I obtain
$$
\sum\limits_{k=0}^{n} \frac{n! (n-k+1)2^{n-k}}{(n-k+1)!(k-1)!}=\sum\limits_{k=0}^{n} \binom{n}{k-1}(n-k+1)2^{n-k}
$$
But it seems strange! As regards 2), I don't know!
 A: HINT for the first problem: $\binom{n}kk2^{n-k}$ is the number of ways to start with $n$ people, choose a team of size $k$, choose one of the team members to be captain, and then choose some subset of the remaining $n-k$ people to be substitutes. Thus, $\sum_{k=0}^n\binom{n}kk2^{n-k}$ is the total number of ways to do this allowing the size of the team to be arbitrary.
You can bet the same effect by choosing one person from the $n$ to be captain, then dividing the remaining $n-1$ people into three groups: the other members of the team, the substitutes, and everyone else.


*

*How many ways are there to choose one person from the $n$ to be captain?  

*How many ways are there to divide $n-1$ people into three sets, any of which can be empty?


Combine the answers to those two questions properly, and you’ll have a closed form for $\sum_{k=0}^n\binom{n}kk2^{n-k}$ as a function of $n$.
There are lots of possible upper bounds for the second summation. One, not especially good, can be obtained by noting that
$$\sum_{k=1}^{n-2}\binom{n}k\frac{k}{n-k}=\sum_{k=1}^{n-2}\binom{n-1}{k-1}\frac{n}{n-k}=n\sum_{k=0}^{n-3}\binom{n-1}k\frac1{n-1-k}$$
and that $\frac1{n-1-k}\le 1$ for $k=0,\ldots,n-1$.
A: For the first one, $(x+2)^n=\sum_{k=0}^n \binom{n}{k}x^k2^{n-k}$. Now differentiate in $x$:
$n(x+2)^{n-1}=\sum_{k=1}^n \binom{n}{k}kx^{k-1}2^{n-k}.$ 
Now plug in $x=1$. 
For the second one, $(x+y)^n=\sum_{k=0}^n \binom{n}{k}x^ky^{n-k}$. Notice that:
$$\frac{(x+y)^n}{y}=\sum_{k=0}^n \binom{n}{k}x^ky^{n-k-1}.$$
Perform partial differentiation in $x$, integrate in $y$, and plug in $x,y=1$. Then correct for any boundary issues in the sum indices. 
A: I prefer the combinatorial approach, but we can do it by manipulation. First note that our sum is 
$$\sum_{k=1}^n k\binom{n}{k}2^{n-k}\tag{1}$$
since the $k=0$ term makes no contribution to the sum.
Then use the fact that $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$ to rewrite (1) as
$$n\sum_{k=1}^n \binom{n-1}{k-1}2^{n-k}.\tag{2}$$
Re-index the sum, replacing $k-1$ by $j$. Then (2) becomes
$$n\sum_{j=1}^{n-1} \binom{n-1}{j}2^{n-1-j}.\tag{3}$$
We recognize the binomial expansion of $(1+2)^{n-1}$, so (3) is equal to 
$$n\cdot3^{n-1}.$$
