i'm doing my homework and I got stuck at the question below and i'm assuming its under random variables because that is what i'm learning at the moment.
for $0 \le p \le 1$ and $n = 2,3,4,....$ determine the value of $\sum_{x=2}^n x(x-1){n\choose x}p^x(1-p)^{n-x} $
I tried to reduce it to this form $ \frac{p^2}{(1-p)^2}\sum_{j=0}^n (n-j+1)(n-j) {n\choose j}p^j(1-p)^{n-j}$ and i'm stuck from here on as im unsure whether my approach is even correct.
Note that
$$x \binom{n}{x} = x \cdot \frac{n!}{x! (n-x)!} = x \cdot \frac{n(n-1)!}{x(x-1)!(n-x)!} = n \cdot \frac{(n-1)!}{(x-1)!((n-1) - (x-1))!} = n \binom{n-1}{x-1}.$$
And we can repeat this:
$$x(x-1)\binom{n}{x} = n(x-1)\binom{n-1}{x-1} = n(n-1)\binom{n-2}{x-2}.$$
So
$$\begin{align} \sum_{x=2}^n x(x-1)\binom{n}{x} p^x (1-p)^{n-x} &= \sum_{x=2}^n n(n-1)\binom{n-2}{x-2} p^x (1-p)^{n-x} \\ &= n(n-1)p^2 \sum_{x=2}^n \binom{n-2}{x-2} p^{x-2} (1-p)^{(n-2) - (x-2)} \\ &= n(n-1)p^2 \sum_{y=0}^{n-2} \binom{n-2}{y} p^y (1-p)^{(n-2) - y} \\ &= n(n-1)p^2 \left(p + (1-p)\right)^{n-2} \\ &= n(n-1)p^2. \end{align}$$ The whole point of this manipulation is to now observe that the last sum corresponds to the sum of all probabilities of outcomes of a binomial random variable with parameters $n-2$ and $p$; by the binomial theorem, it has value $1$.
The relationship we established at the beginning is easily generalized:
$$x(x-1)\cdots(x-k+1) \binom{n}{x} = n(n-1)\cdots(n-k+1) \binom{n-k}{x-k},$$ where $1 \le k \le n$. Equivalently,
$$\binom{x}{k} \binom{n}{x} = \binom{n}{k} \binom{n-k}{x-k}.$$
This in turn implies for $X \sim \operatorname{Binomial}(n,p)$ and $k \in \{1, 2, \ldots, n\}$,
$$\operatorname{E}\left[\binom{X}{k}\right] = \binom{n}{k} p^k.$$
As we have seen, your question corresponds to the special case $k = 2$:
$$\operatorname{E}[X(X-1)] = 2\operatorname{E}\left[\binom{X}{2}\right] = 2\binom{n}{2} p^2 = 2 \frac{n(n-1)}{2} p^2 = n(n-1)p^2.$$