How $\frac{1}{n}\sum_{i=1}^n X_i^2 - \bar X^2 = \frac{\sum_{i=1}^n (X_i - \bar X)^2}{n}$ How 
$\frac{1}{n}\sum_{i=1}^n X_i^2 - \bar X^2 = \frac{\sum_{i=1}^n (X_i - \bar X)^2}{n}$ 
i have tried to do that by the following procedure:
$\frac{1}{n}\sum_{i=1}^n X_i^2 - \bar X^2$
=$\frac{1}{n}(\sum_{i=1}^n X_i^2 - n\bar X^2)$
=$\frac{1}{n}(\sum_{i=1}^n X_i^2 - \sum_{i=1}^n\bar X^2)$
=$\frac{1}{n} \sum_{i=1}^n (X_i^2 - \bar X^2)$
Then i have stumbled.
 A: I think it is cleaner to expand the right-hand side. We have
$$(X_i-\bar{X})^2=X_i^2-2X_i\bar{X}+(\bar{X})^2.$$
Sum over all $i$, noting that $2\sum X_i\bar{X}=2n\bar{X}\bar{X}=2n(\bar{X})^2$ and $\sum (\bar{X})^2=n(\bar{X})^2$. 
There is some cancellation. Now divide by $n$ and we get the left-hand side. 
A: Expand the square and use that $\sum\limits_{k=1}^nX_k=n\bar{X}$:
$$
\begin{align}
\frac1n\sum_{k=1}^n(X_k-\bar{X})^2
&=\frac1n\sum_{k=1}^n(X_k^2-2X_k\bar{X}+\bar{X}^2)\\
&=\frac1n\left(\sum_{k=1}^nX_k^2-2n\bar{X}\bar{X}+n\bar{X}^2\right)\\
&=\frac1n\sum_{k=1}^nX_k^2-\bar{X}^2\\
\end{align}
$$
A: Since you tagged it under statistics, note that the LHS is simply $E[X^2] - E[X]^2$, and the RHS is $E[(X-\bar{X})^2]$. Both of these are ways to write Variance of $X$, hence they are equal.
A: \begin{align}
\sum_{i=1}^n (X_i-\bar X)^2 & = \sum_{i=1}^n (X_i^2-2X_i\bar X + (\bar X)^2) \\[10pt]
& = \left(\sum_{i=1}^n (X_i^2)\right) - \left( \sum_{i=1}^n 2X_i \bar X \right) + \left(\sum_{i=1}^n ((\bar X)^2) \right).
\end{align}
In the second term, the factor $2\bar X$ does not change as $i$ goes from $1$ to $n$, so it can be pulled out, getting
$$
2\bar X\sum_{i=1}^n X_i.
$$
Then we can say that $\displaystyle\sum_{i=1}^n X_i = n\bar X$.  The second term is then $2n(\bar X)^2$.
In the third term, the sum is $(\bar X)^2 +\cdots+(\bar X)^2$.  This is just $n(\bar X)^2$.
Now we have
$$
\left(\sum_{i=1}^n (X_i^2)\right) - 2n(\bar X)^2 + n(\bar X)^2.
$$
The last two terms can be collected into one term.
