The problem asks us to calculate the sum $\sum_{n=1}^{6} (n^2 - 1)$.

1. Problem Description

The problem asks us to calculate the sum $\sum_{n=1}^{6} (n^2 - 1)$.

2. Solution Steps

We need to evaluate the sum $\sum_{n=1}^{6} (n^2 - 1)$. We can split this sum into two separate sums:

$\sum_{n=1}^{6} (n^2 - 1) = \sum_{n=1}^{6} n^2 - \sum_{n=1}^{6} 1$.

We know the formula for the sum of the first $n$ squares:

$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.

In our case, $n = 6$, so

$\sum_{n=1}^{6} n^2 = \frac{6(6+1)(2(6)+1)}{6} = \frac{6(7)(13)}{6} = 7(13) = 91$.

The sum of $1$ from $n=1$ to $n=6$ is simply $6$, so

$\sum_{n=1}^{6} 1 = 6$.

Therefore,

$\sum_{n=1}^{6} (n^2 - 1) = \sum_{n=1}^{6} n^2 - \sum_{n=1}^{6} 1 = 91 - 6 = 85$.

3. Final Answer

85