Proof by induction of $\sum_{k=2}^n (k-1)(k)\binom{n}{k} = n(n-1)2^{n-2}$ I've been struggling with this sum for an while, plugging $n+1$ instead of $n$, knowing that $\binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}$  and after some manipulation  I've found this sum.
$$2\sum_{k=1}^{n} k^2\binom{n}{k}$$
I couldn't see any way I could get out of here and I don't know how to start this proof without the property of the sum of binomial coefficients.
Thanks in advance.
 A: I'm not sure if you want any proof or specifically a proof by induction, but here is a proof. 
Recall that $\sum_{k=0}^n\binom{n}k x^k = (1+x)^n$ by the binomial theorem. Then, differentiating both sides twice with respect to $x$, we find that
$$
\sum_{k=2}^n k(k-1)\binom{n}k x^k = n(n-1)(1+x)^{n-2}. 
$$
Then, plugging in $x=1$ will give you the desired result. 
A: You'll have to forgive me if I've misunderstood something but assuming $$\sum^{n}_{k=2}(k-1)(k){{n}\choose{k}}=n(n-1)2^{n-2}$$ is the requirement for the inductive proof then here I go.
Using the identity 
$$i{{n}\choose{i}}=n{{n-1}\choose{i-1}}$$ we have
$$\sum^{n}_{k=2}(k-1)(k){{n}\choose{k}}=n(n-1)2^{n-2} \Rightarrow$$
$$n\sum^{n}_{k=2}(k-1){{n-1}\choose{k-1}}=n(n-1)2^{n-2}\Rightarrow$$
$$n(n-1)\sum^{n}_{k=2}{{n-2}\choose{k-2}}=n(n-1)2^{n-2}$$
This means we just need to prove inductively that 
$$\sum^{n}_{k=2}{{n-2}\choose{k-2}}=2^{n-2}$$ which is apparently a rewriting of
$$(x+y)^{n}=\sum_{k=0}^{n}{{n}\choose{k}}x^{n-k}y^{k}$$ where $x$ and $y$ are both set to $1$.
A: Here is a combinatorial proof.  
We count committees of $k$ people with a chairperson and a vice-chairperson selected from a group with $n$ people in two ways.
To form a committee of size $k$ with a chairperson and a vice-chairperson, we can first choose $k$ of the $n$ people to serve on the committee, then select one of those $k$ people to serve as the chairperson, and one of the remaining $k - 1$ people to serve as the vice-chairperson.  The number of ways to do this is 
$$k(k - 1)\binom{n}{k}$$
The summation on the left-hand side counts all such committees.
Alternatively, we can select one of the $n$ people to be the chairperson, one of the remaining $n - 1$ people to be the vice-chairperson, and then choose a subset of the remaining $n - 2$ people to serve on the committee with them.  Since there are $2^{n - 2}$ subsets of a set with $n - 2$ elements, the number of ways this can be done is 
$$n(n - 1)2^{n - 2}$$
Since both sides of the equation count the same set of committees, we may conclude that 
$$\sum_{k = 2}^{n} k(k - 1)\binom{n}{k} = n(n - 1)2^{n - 2}$$
A: Applying Pascal's identity gives you
$$ \sum_k k(k - 1) \binom{n + 1}{k} = \sum_k k(k - 1) \binom{n}{k} + \sum_k k(k - 1) \binom{n}{k - 1}. $$
If we call the sum $f(n)$ then this says that
$$ f(n + 1) = f(n) + \text{ a sum that looks very much like } f(n). \tag{1}$$
We can deal with the second sum by changing variables:
$$ \sum_k k(k - 1) \binom{n}{k - 1} = \sum_k (k + 1)k \binom{n}{k} = f(n) + 2\sum_k k \binom{n}{k}. \tag{2}$$
Let $g(n) = \sum_k k \binom{n}{k}$. Then similarly we find that
$$ g(n + 1) = g(n) + \sum_k (k + 1) \binom{n}{k} = 2g(n) + \sum_k \binom{n}{k} = 2g(n) + 2^n. $$
Therefore
\begin{align}
g(n + 1) &= 2g(n) + 2^n \\
&= 2\left[2g(n - 1) + 2^{n - 1} \right] + 2^n \\
&= 4g(n - 1) + 2 \cdot 2^n \\
&= 8g(n - 2) + 3 \cdot 2^n \\
\end{align}
and by induction
$$g(n + 1) = 2^{n + 1}g(0) + (n + 1)2^n = (n + 1)2^n. \tag{3}$$
Substituting $(2)$ and $(3)$ into $(1)$ gives us
$$ f(n + 1) = 2f(n) + 2g(n) = 2f(n) + n2^n. $$
Hence induction gives
$$ f(n + 1) = n2^n + (n - 1)2^n + (n - 2)2^n + \dots + 1 \cdot 2^n = n(n + 1)2^{n - 1}. $$
A: A proof by induction seems to be somewhat cumbersome. The essence in the answer of @Vaas is the relationship
\begin{align*}
\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}=\frac{n(n-1)}{k(k-1)}\binom{n-2}{k-2}\tag{1}
\end{align*}

Using (1) we obtain for $n\geq 2$:
  \begin{align*}
\color{blue}{\sum_{k=2}^nk(k-1)\binom{n}{k}}&=n(n-1)\sum_{k=2}^n\binom{n-2}{k-2}\\
&=n(n-1)\sum_{k=0}^{n-2}\binom{n-2}{k}\tag{2}\\
&\color{blue}{=n(n-1)2^{n-2}}
\end{align*}

In (2) we shift the index $k$ to start from $k=0$ and are ready to apply the binomial theorem.
