1. The problem is to verify or understand the formula for the sum of squares of the first $n$ natural numbers: $$12 + 22 + 32 + \cdots + n^2 = 6n(n+1)(2n+1)$$ 2. The correct formula for the sum of squares is actually: $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$ 3. The given formula seems to have a formatting issue. It should be interpreted as the sum of squares equals $\frac{n(n+1)(2n+1)}{6}$, not multiplied by 6. 4. Let's verify the formula for a small value, say $n=3$: $$1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14$$ Using the formula: $$\frac{3 \times 4 \times 7}{6} = \frac{84}{6} = 14$$ 5. This confirms the formula is: $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$ 6. Therefore, the sum of squares from 1 to $n$ is given by this formula, which is a standard result in algebra. Final answer: $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$