What is the sum of the squares of the digits from $1$ to $9$?

### Concept & Formula The sum of the squares of the first $n$ natural numbers follows a specific algebraic series formula. Instead of manually squaring and adding each digit, applying this formula yields the result directly and safely. The standard formula is: $$\Sigma n^2 = \frac{n(n+1)(2n+1)}{6}$$ ### Step-by-Step Solution **Given:** * We need to find the sum of squares of digits from $1$ to $9$. * $n = 9$ **Calculation / Deduction:** Substitute $n = 9$ into the sum of squares formula: $$\Sigma 9^2 = \frac{9(9+1)(2(9)+1)}{6}$$ $$\Sigma 9^2 = \frac{9 \times 10 \times 19}{6}$$ Simplify the fraction by dividing $9$ and $6$ by $3$: $$\Sigma 9^2 = \frac{3 \times 10 \times 19}{2}$$ Divide $10$ by $2$: $$\Sigma 9^2 = 3 \times 5 \times 19$$ $$\Sigma 9^2 = 15 \times 19$$ Calculate the final product: $$15 \times 19 = 15 \times (20 - 1) = 300 - 15 = 285$$ ### Exam Strategy & Shortcut For competitive exams, memorizing the sums of squares up to $10$ is a massive time-saver. The sum of squares up to $10$ is famously $385$. To find the sum up to $9$, simply subtract the square of $10$: $385 - 100 = 285$. This guarantees an instantaneous answer. ### Common Pitfall A common error is confusing the sum of squares formula with the square of the sum formula $(\frac{n(n+1)}{2})^2$. Using the wrong formula for $n=9$ would give $45^2 = 2025$, leading to confusion and wasted time when it doesn't match the options. ### Final Answer Therefore, the correct answer is 285.