Simplify $\sum_{i} i\, {{n}\choose{ i}}$ I guess it would be $\sum_{i} i\, {{n}\choose{ i}} = \sum_{i}n{{n-1}\choose{ i-1}} = n 2^{n-1}$. What is combinational interpretation of $\sum_{i} i\, {{n}\choose{ i}} = n 2^{n-1}$?
 A: I have $n$ different doughnuts in front of me, and want to choose $1$ to eat right away, and choose  some (possibly none) of the rest to put in a bag for lunch, saving the rest for dinner. How many ways can I do this?
One way of counting: There are $n$ ways to choose the one to eat right away, and for each of these ways there are $2^{n-1}$ ways to choose the subset of the remaining $n-1$ to be put in a bag for lunch. This gives a total of $n2^{n-1}$ ways.
Another way of counting: I decide on the total number $i$ of doughnuts to be used for now and lunch combined. So $i$ can be anything from $1$ to $n$. 
There are  $\binom{n}{i}$ ways to choose these $i$ doughnuts. Once they have been chosen, there are $i$ ways to choose from them the one  to eat right away, putting the remaining $i-1$ in a bag for lunch. Thus for any $i$, this choosing can be done in $i\binom{n}{i}$ ways. Add up over all $i$.
Remark: Or else we have $n$ kids in a class. We want to choose $1$ of them to receive a gold medal, some, possibly none, to receive a silver medal, and the rest to receive a plastic medal.
A: Another way
(definitely not original):
If
$s_n
= \sum_{i} i\, {{n}\choose{ i}}
$,
then
$\begin{array}\\
s_n
&= \sum_{i=0}^n i\, {{n}\choose{ i}}\\
&= \sum_{i=0}^n (n-i)\, {{n}\choose{n- i}}\\
&= \sum_{i=0}^n n\, {{n}\choose{i}}-\sum_{i=0}^n i\, {{n}\choose{ i}}\\
&= n2^n-s_n\\
\text{so}\\
s_n
&= n2^{n-1}\\
\end{array}
$
