Algebric proof for the identity $n(n-1)2^{n-2}=\sum_{k=1}^n {k(k-1) {n \choose k}}$

Solution 1:

HINT:

For $k\ge2,$

$$k(k-1)\binom nk=k(k-1)n(n-1)\frac{(n-2)!}{k(k-1)\cdot (k-2)! \{(n-2)-(k-2)\}!}$$

$$=n(n-1)\binom{n-2}{k-2}$$

Solution 2:

The $n(n-1)$ and $k(k-1)$ are a signal that differentiating twice should be at hand.

So start with $$ (1+X)^n=\sum_k\binom nkX^k, $$ differentiate twice with respect to $X$ to get$$ n(n-1)(1+X)^{n-2}=\sum_k\binom nk k(k-1)X^k, $$ and finally set $X:=1$.

Algebric proof for the identity $n(n-1)2^{n-2}=\sum_{k=1}^n {k(k-1) {n \choose k}}$

Solution 1:

Solution 2:

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