Sum Of Special Series

The sum of the first n perfect squares has the closed form n(n+1)(2n+1)/6, a special-series identity frequently invoked in JEE Advanced telescoping and counting problems. Substituting n = 10 gives 10 × 11 × 21 / 6. Compute the numerator: 10 × 11 = 110, and 110 × 21 = 2310. Dividing by 6 yields 2310/6 = 385. Option 330 corresponds to the sum of the first ten triangular-like terms using a wrong factor. Option 285 equals the sum of squares only up to n = 9, missing the last term 100. Option 405 overshoots by misapplying (2n+1) as (2n+3). Hence the required sum is 385. Plausibility check: the average square value over 1 to 100 is 38.5, and ten terms times that average gives 385, while the answer also exceeds 9^2 + previous total, confirming the cumulative growth is reasonable.