This is problem 2.7.3.1 from "Theory of Point Estimation." We define the statistic $T = \sum X_i$ to be the number of successes in a binomial distribution. We're trying to show that: $\sum (X_i - \bar{X})^2 = \frac{T(n-T)}{n}$
So I was trying to work it out:
$$ \frac{T(n-T)}{n} = \frac{\sum X_i(n-\sum X_i)}{n} = \sum X_i - \frac{1}{n}(\sum X_i)^2 = \sum X_i - \left(\frac{\sum X_i}{n}\right)\sum X_i = \sum X_i - \bar{X}\sum X_i = \sum X_i \left(1- \bar{X} \right)$$
I don't feel like I'm getting any closer to the solution.
The first key observation is that if $T = \sum X_i$, then $T = n\bar X$. Therefore, $$\begin{align*} \sum (X_i - \bar X)^2 &= \sum (X_i - T/n)^2 \\ &= \sum \left(X_i^2 - \frac{2T}{n} X_i + \frac{T^2}{n^2} \right) \\ &= \sum X_i^2 - \frac{2T}{n} T + \frac{T^2}{n} \\ &= \sum X_i^2 - \frac{T^2}{n}.\end{align*}$$ The second key observation is that $X_i \in \{0,1\}$, as these are independent and identically distributed Bernoulli variables. Therefore, $X_i^2 = X_i$ for each $i = 1, \ldots, n$, since if $X_i = 0$, then $X_i^2 = 0$, and if $X_i = 1$, then $X_i^2 = 1$. Consequently, $$\sum X_i^2 = \sum X_i = T,$$ and we have $$\sum (X_i - \bar X)^2 = T - \frac{T^2}{n} = \frac{T(n-T)}{n},$$ as claimed.