Sum Of Special Series
Evaluate the value of the finite sum 1^2 + 2^2 + 3^2 + ... + 10^2 using the standard formula for the sum of squares.
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About This Question
- Subject
- mathematics
- Chapter
- sequence and series
- Topic
- sum of special series
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
385
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.
This easy difficulty mathematics question is from the chapter sequence and series, covering the topic of sum of special series. It appeared in the 2025 exam.
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