Show that $\Sigma_{i=1}^{n}(x_i - \bar x_n )^2$ can be calculated with a recursion Specifically with the recursion
$$\Sigma_{i=1}^{k+1}(x_i - \bar x_{k+1} )^2 = \Sigma_{i=1}^{k}(x_i - \bar x_k )^2 + \frac{k}{k+1}(x_{k+1} -\bar x_k)^2 $$
for $k = 1...n-1$. 
I know that 
$$\bar x_{k+1} = \frac{k}{k+1}\bar x_k + \frac{k}{k+1}\bar x_{k+1}$$
but after that I'm not sure. I showed that it's true for $k = 2$ but after that I'm stuck. I tried opening up the right hand side but just ended up in a morass of equations.
 A: Indeed there are a lot of equations. To simplify call the sum  $Q_n,$ i.e. we have
\begin{align*}
Q_n &= \sum_{i=1}^n(x_i-\bar{x}_n)^2\\
&=\sum_{i=1}^n(x_i^2-2 x_i\bar{x}_n +\bar{x}_n^2)\\
&= \sum_{i=1}^n x_i^2- 2\bar{x}_n \sum_{i=1}^nx_i + \sum_{i=1}^n\bar{x}_n^2\\
&= \sum_{i=1}^n x_i^2- 2\bar{x}_n (n\bar{x}_n) + n\bar{x}_n^2\\
&= \sum_{i=1}^n x_i^2- n\bar{x}_n^2
\end{align*}
then start computing the difference
\begin{align*}
Q_n- Q_{n-1} &= \sum_{i=1}^n x_i^2- n\bar{x}_n^2 - \sum_{i=1}^{n-1} x_i^2+(n-1)\bar{x}_{n-1}^2\\
&= x_n^2 - n\bar{x}_n^2 + (n-1)\bar{x}_{n-1}^2\\
&= x_n^2 - \bar{x}_{n-1}^2 + n(\bar{x}_{n-1}^2-\bar{x}_{n}^2)\\
&=x_n^2 - \bar{x}_{n-1}^2 + n(\bar{x}_{n-1}-\bar{x}_{n})(\bar{x}_{n-1}+\bar{x}_{n}).
\end{align*}
Now use $n(\bar{x}_{n-1}-\bar{x}_{n})=\bar{x}_{n-1}-x_n$ which follows from
$$
n\bar{x}_n = \sum_{i=1}^n x_i = \sum_{i=1}^{n-1} x_i + x_n = (n-1)\bar{x}_{n-1} + x_n
$$
and get  the difference:
\begin{align*}
Q_n- Q_{n-1} &= x_n^2 - \bar{x}_{n-1}^2 + (\bar{x}_{n-1}-x_n)(\bar{x}_{n-1}+\bar{x}_{n})\\
&=x_n^2 - \bar{x}_{n-1}^2 +
\bar{x}_{n-1}^2 + \bar{x}_{n-1}\bar{x}_{n}-x_n\bar{x}_{n-1}-x_n\bar{x}_{n}\\
&=x_n^2  + \bar{x}_{n-1}\bar{x}_{n}-x_n\bar{x}_{n-1}-x_n\bar{x}_{n}\\
&=(x_n-\bar{x}_{n-1})(x_n-\bar{x}_n)
\end{align*}
With
\begin{align*}
n(\bar{x}_n-x_n) &= \sum_{i=1}^n x_i - n x_n\\
&=\sum_{i=1}^{n-1} x_i + x_n - nx_n\\
&=\sum_{i=1}^{n-1} x_i - (n-1) x_n \\
&= (n-1)\bar{x}_{n-1} - (n-1) x_n \\
&= (n-1)(\bar{x}_{n-1} - x_n) \\
\end{align*}
we have
$$
x_n -\bar{x}_n = \left(\frac{n-1}{n}\right)\left(x_{n} -\bar{x}_{n-1}\right)
$$
and finally

$$
Q_n -Q_{n-1} = \left(\frac{n-1}{n}\right)\left(x_{n} -\bar{x}_{n-1}\right)^2.
$$

