Answer
The number of comparisons is $\frac{n(n+1)-2}{2}$
Work Step by Step
Looking at the insertion sort algorithm, we can deduce a few things:
We have $n-1$ loops and for each loop we do a certain number of comparisons. In our case, the values are arranged in ascending order.
Looking at the inner loop, it is easy to note that for some loop iteration $i$, we have to compare $a_i$ against all the values before it until we reach $a_i$ again; hence, we have $i$ comparisons.
We can sum those deductions in one equation:
Number of comparisons $:= \sum_{i=2}^n{i} = 2+3+... n = \left(\sum_{i=0}^{n}i\right) - 1.$
The standard result is (from chapter 2:) $\sum_{i=0}^ni=\frac{n(n+1)}{2}$ .
Applying this to the formula above, we get the number of comparisons to be $\frac{n(n+1)}{2} - 1 = \frac{n(n+1)-2}{2}$
