1. The problem is to find the formula for the sum of the squares of the first $n$ natural numbers, i.e., calculate $\sum_{i=1}^n i^2$. 2. The formula for the sum of squares is: $$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$ This formula allows us to find the sum quickly without adding each square individually. 3. Let's understand why this formula works by verifying it 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. 4. To use the formula for any $n$, simply substitute the value of $n$ into the formula and simplify. 5. For example, for $n=5$: $$\sum_{i=1}^5 i^2 = \frac{5 \times 6 \times 11}{6}$$ We can cancel the 6 in numerator and denominator: $$= \frac{5 \times \cancel{6} \times 11}{\cancel{6}} = 5 \times 11 = 55$$ 6. So, the sum of squares from 1 to 5 is 55. This formula is very useful in algebra and calculus for summation problems involving squares.