The problem asks us to evaluate the sum $\sum_{n=4}^{9} (n-1)$.

1. Problem Description

The problem asks us to evaluate the sum $\sum_{n=4}^{9} (n-1)$.

2. Solution Steps

We can expand the summation as follows:

$\sum_{n=4}^{9} (n-1) = (4-1) + (5-1) + (6-1) + (7-1) + (8-1) + (9-1)$

$\sum_{n=4}^{9} (n-1) = 3 + 4 + 5 + 6 + 7 + 8$

Now, we add the numbers together:

$3 + 4 + 5 + 6 + 7 + 8 = 7 + 5 + 6 + 7 + 8 = 12 + 6 + 7 + 8 = 18 + 7 + 8 = 25 + 8 = 33$

Alternatively, we can use the formula for the sum of an arithmetic series. First, let $k = n-1$. Then, when $n=4$, $k = 4-1 = 3$, and when $n=9$, $k=9-1=8$. Thus we have

$\sum_{n=4}^{9} (n-1) = \sum_{k=3}^{8} k$

We can calculate $\sum_{k=1}^{8} k$ and $\sum_{k=1}^{2} k$, and subtract the second from the first.

The sum of the first $n$ integers is given by

$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

So $\sum_{k=1}^{8} k = \frac{8(8+1)}{2} = \frac{8(9)}{2} = \frac{72}{2} = 36$

And $\sum_{k=1}^{2} k = \frac{2(2+1)}{2} = \frac{2(3)}{2} = 3$

So $\sum_{k=3}^{8} k = \sum_{k=1}^{8} k - \sum_{k=1}^{2} k = 36 - 3 = 33$

3. Final Answer

33