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!

Calculate the sum of squares from 1^2 to 31^2 using the sum of squares formula. This detailed step-by-step solution demonstrates how to apply the formula S = n(n+1)(2n+1)/6 to get the sum 10,416. Suitable for students in Grades 8-10.

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