Question 1073041
So think of an arithmetic series that starts at 1 and advances by 1. 
The sum of that series would be,
{{{S[n]=(n/2)(1+n)}}}
So then the average would be the sum divided by n.
{{{A[n]=(n+1)/2}}}
So then to find the standard deviation, find the sum,
{{{sum((k-(n+1)/2)^2,k=1,n)=(n(n^2-1))/12}}}
Now divide that sum by n and take the square root,
{{{sigma=sqrt(n(n^2-1))/(12n)}}}
{{{sigma=sqrt((n^2-1))/12)}}}