To solve $1^2 + 2^2 + 3^2 + \cdots + 31^2$, we can use the formula for the sum of squares of the first $n$ natural numbers:

$S = 1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$

In this case, $n = 31$. Let's calculate it step by step.

$S = \frac{31(31+1)(2 \cdot 31 + 1)}{6}$

$S = \frac{31 \cdot 32 \cdot 63}{6}$

$S = \frac{31 \cdot 32 \cdot 63}{6} = \frac{62496}{6} = 10416$

Thus, the sum $1^2 + 2^2 + \cdots + 31^2$ equals 10,416.

Would you like further details or explanations on any part of this? Here are some related questions:

1. How can you derive the formula for the sum of squares?
2. What is the sum of squares up to $n = 50$?
3. Can this formula be used to calculate sums for non-consecutive numbers?
4. What is the sum of the first $n$ cubes, and how does it compare to the sum of squares?
5. What is the average value of the squares in this sum?

Tip: Always break complex sums into formula-based solutions for efficiency!