<s>[INST] 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.
 [/INST] 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
$$ f(n + 1) = n2^n + (n - 1)2^n + (n - 2)2^n + \dots + 1 \cdot 2^n = n(n + 1)2^{n - 1}. $$
</s>