Need to solve a sum

Question

I need to solve a sum: $$\sum_{k=0}^{n} k(k-1)C_{n}^{k} $$

How to start and what formulas or theorems I should use?

Answer 1

HINT

Recall that by binomial theorem

$$(1+x)^n = \sum_{k=0}^n {n \choose k} x^k$$

then we can use derivative and set a convenient value for $x$.

Answer 2

To begin with, $x^k$ is a continuous function. So we interchange infinite sums and derivatives involving this function. Now write $$ \sum_{k=1}^{n} \frac{\partial{^2}}{\partial{x^2}}(x^k ) \times \binom{n}{k} = \frac{\partial{^2}}{\partial{x^2}} \sum_{k=1}^{n}\binom{n}{k}x^k $$ and follow @gimusi's solution

EDIT: set LHS for this expression to $G(x)$ and to get the final result set $x=1$

Answer 3

Hint:

$\displaystyle\sum_{k=0}^n\binom nk x^k=(1+x)^n$, so $$\bigl((1+x)^n\big)''=\sum_{k=0}^n\binom nk k(k-1)x^{k-2}.$$

Answer 4

Question: On how many ways can we choose a group in a set of $n$ people and then president and then vicepresident?

Well we can first a group of $k$ people, that is ${n\choose k}$, for every $k\leq n$, and then president among them, so we have $k$ choises and then $k-1$ choises for V.P. Suming for all $k$ we get:

$$\sum_{k=0}^{n} k(k-1)C_{n}^{k} $$

On the other hand we can first choose a president among all people, so we have $n$ posibilities and then V.P. for who we have $n-1$ choises and then we choose any set in set of $n-2$ people, for that we have $2^{n-2}$ choises, so:

$$n(n-1)2^{n-2}$$ and this is the answer to your question.