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The image shows the formula for the sum of squares of the first \(n\) natural numbers:

\[
\sum_{i=1}^{n} i^2 = 1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}
\]

This equation expresses the sum of squares as a closed-form formula, which simplifies calculating the sum of squares for large \(n\) without manually computing each square. Here's a breakdown of the formula:

- \( n \): The number of terms (natural numbers) for which you want to calculate the sum of squares.
- \( \sum_{i=1}^{n} i^2 \): The summation notation representing the sum of squares of all integers from 1 to \(n\).
- The right-hand side formula: \(\frac{n(n+1)(2n+1)}{6}\) efficiently computes this sum.

For example:
- If \( n = 3 \), the sum of squares is \( 1^2 + 2^2 + 3^2 = 1 + 4 + 

The formula for the sum of squares of the first n natural numbers, as shown in the image.

Discover how to calculate the sum of squares of the first n natural numbers using the closed-form formula: \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \). This solution helps simplify the calculation for large n, making it ideal for algebra students in Grades 10-12.

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