Proving an identity algebraically Show that $$\sum_{i=0}^{n} \binom{n}{i} i (i-1)=n(n-1)2^{n-2}$$
knowing that $$\sum_{i=0}^{n}\binom{n}{i}i=n2^{n-1}$$
I ended up after a while with
$$\sum_{i=0}^{n}\frac{n-1}{2}\binom{n}{i}i=n(n-1)2^{n-2}$$
How can I transform of the $\frac{n-1}{2}$?
 A: Note that $ \dbinom{n}{i} i(i-1) = \frac{n!}{(n-i)! (i - 2)!}= n(n-1) \dbinom{n-2}{i} $. 
Because $ \dbinom{n}{-2} = \dbinom{n}{-1} = 0 $, $$\sum_{i = 0}^n n(n-1)\dbinom{n-2}{i-2} = n(n-1) \sum_{i = 0}^{n-2} \dbinom{n-2}{i} = n(n-1)2^{n-2} $$
This only requires the knowledge that $ \sum_{i = 0}^n \dbinom{n}{i} = 2^n $ which is immediately obvious from the Binomial Theorem applied to $ (x + y)^n $ when $ x = y = 1 $. 
If you wanted to explicitly use the given sum, you would consider $ \displaystyle \sum_{i = 0}^n \dbinom{n}{i} i^2 $. From there, $ \dbinom{n}{i} i^2 = n\dbinom{n-1}{i - 1}i $. Then shift the index up by $ 1 $ and you will once again have a linear function of $ i $ and $ \dbinom{n}{i} $. 
A: $\text{Consider}$ $$\sum_{i=0}^n \dbinom{n}i x^i = (1+x)^n \tag{$\star$}$$
$\text{Differentiating $(\star)$, we get that}$
$$\sum_{i=0}^n \dbinom{n}i i x^{i-1} = n(1+x)^{n-1} \tag{$\perp$}$$
$\text{Differentiating $(\perp)$, we get that}$
$$\boxed{\color{red}{\displaystyle \sum_{i=0}^n \dbinom{n}i i(i-1) x^{i-2} = n(n-1)(1+x)^{n-2}}}$$
$\text{Now, set $x=1$ to get the answer, i.e.,}$
$$\boxed{\color{blue}{\displaystyle \sum_{i=0}^n \dbinom{n}i i(i-1) = n(n-1)2^{n-2}}}$$
A: You can simply pull the constant factor of $\frac{n-1}2$ outside the summation: that’s just the distributive law. Alternatively, you start at the beginning and use the identity
$$\binom{n}ii=n\binom{n-1}{i-1}\;,$$
which you can verify either combinatorially or by expanding into factorials.
Then
$$\binom{n}i(i-1)=n\binom{n-1}{i-1}(i-1)=n(n-1)\binom{n-2}{i-2}\;,$$
and you can pull the factor of $n(n-1)$ out of the summation. Then you need only know that $\sum_{k=0}^m\binom{m}k=2^m$.
