The problem asks to evaluate the sum $\sum_{n=1}^{6} n^2$.

1. Problem Description

The problem asks to evaluate the sum $\sum_{n=1}^{6} n^2$.

2. Solution Steps

We need to calculate the sum of the squares of the first 6 natural numbers.

We can use 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, $n = 6$. Therefore,

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

$= \frac{6(7)(12+1)}{6}$

$= \frac{6(7)(13)}{6}$

$= 7 \times 13$

$= 91$

3. Final Answer

91