Formula for standard deviation So I am studying for a final exam and I know, or thought I knew, the formula for standard deviation. I quadruple checked in my notes and the slides for the lecture and see that the formula for the standard deviation when $\sigma$ is unknown is:

$\sqrt((\Sigma d^2 - nd^-) /  n - 1)$ 

But for some reason on the final review solution sheet our professor somehow and some reason replaced $n$ in the numerator with $1/n$ and I have no clue why. So the formula is now:

$\sqrt((\Sigma d^2 - (1/n)d^-) / n - 1)$

My question is, how come we are able to use $1/n$ instead of $n$ and on the test, how can I tell to use $1/n$ as compared to $n$?

Thanks
Yes your professor has a typo.  The standard deviation for paired samples is usually written as follows:

$$\sigma = \sqrt{\frac{\sum_{i=1}^n(d_i-\bar d)^2}{n-1}}$$

What you professor is doing is using the trick (or arithmetic if you want to think of it that way) that we can rewrite the numerator as

$$\sum_{i=1}^n(d_i-\bar d)^2 = \sum_{i=1}^nd_i^2-n\bar d^2$$

And thus, the formula for the standard deviation should be 

$$\sigma = \sqrt{\frac{\sum_{i=1}^n(d_i-\bar d)^2}{n-1}}= \sqrt{\frac{\sum_{i=1}^n(d_i)^2-n\bar d^2}{n-1}}$$

And finally, it should be $n$ and not $\frac{1}{n}$.
Your formulas are wrong as $\bar d$ needs to be squared.

Now, you can indeed meet valid formulas with $n$ and $\dfrac1n$, because

$$\color{blue}n{\bar d}^2=n\left(\frac1n\sum d_i\right)^2=\color{blue}{\frac1n}\left(\sum d_i\right)^2.$$

Confusion can arise from this.