1. **Stating the problem:** We want to understand and verify the formula for the sum of squares of the first $n$ natural numbers, which is given as: $$P(n) = 1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$ 2. **Formula used:** The sum of squares formula is a well-known closed form expression that allows us to calculate the sum without adding each square individually. 3. **Explanation:** The formula states that the sum of the squares from 1 to $n$ equals the product of $n$, $n+1$, and $2n+1$, divided by 6. 4. **Intermediate work:** Let's verify this formula for a small value, say $n=3$: $$1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14$$ Using the formula: $$\frac{3 \times 4 \times 7}{6} = \frac{84}{6} = 14$$ This matches the direct sum. 5. **Learner-friendly explanation:** Instead of adding each square one by one, this formula gives a quick way to find the total sum. It works for any positive integer $n$. **Final answer:** $$P(n) = \frac{n(n+1)(2n+1)}{6}$$