Skip to content

Sum Of Special Series

Easymathematics

Evaluate the value of the finite sum 1^2 + 2^2 + 3^2 + ... + 10^2 using the standard formula for the sum of squares.

Select the correct option:

🔒 Solution Hidden from View

Submit your answer to unlock the detailed step-by-step solution.

More sequence and series Practice Questions

MediumJEE_ADVANCED

The general term of a series is given by T_k = k(k+2); evaluate the sum of the first n terms of this...

View Solution→
MediumJEE_ADVANCED

The sum of the first n terms of a certain sequence equals S_n = 3n^2 + 5n for every positive integer...

View Solution→
HardJEE_ADVANCED

Assertion: the series with terms 1/n diverges; Reason: every series whose terms tend to zero must co...

View Solution→
HardJEE_ADVANCED

If the numbers a, b, c are simultaneously in arithmetic progression and in geometric progression wit...

View Solution→
HardJEE_ADVANCED

Determine the exact value of the convergent infinite sum whose kth term is k/2^k, taken over all pos...

View Solution→
HardJEE_ADVANCED

If positive numbers x, y, z are in geometric progression, then the logarithms log x, log y, log z to...

View Solution→
EasyJEE_ADVANCED

Consider the infinite geometric series 9 - 6 + 4 - 8/3 + ... that converges; what is its exact sum t...

View Solution→
MediumJEE_ADVANCED

Given that the fourth term of a harmonic progression is 1/12 and its ninth term is 1/27, find the tw...

View Solution→
MediumJEE_ADVANCED

Determine the sum to \infty of the arithmetico-geometric series 1 + 2x + 3x^2 + 4x^3 + ... when the...

View Solution→
MediumJEE_ADVANCED

Find the sum of the telescoping series whose general term is 1/(k(k+1)) summed from k equal to 1 up ...

View Solution→
EasyJEE_ADVANCED

If three arithmetic means are inserted between the numbers 4 and 28 to form an arithmetic progressio...

View Solution→
MediumJEE_ADVANCED

For positive reals a, b, c whose product abc equals 27, what is the minimum possible value of the su...

View Solution→

Browse all sequence and series questions →

About This Question

Subject
mathematics
Chapter
sequence and series
Topic
sum of special series
Difficulty
Easy
Year
2025
Tags
advanced-calculus-drillsum-of-squaresspecial-seriesclosed-formfinite-sum

Solution

Correct Answer:

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.

Looking for more practice? Explore all mathematics questions or browse sequence and series questions on RankGuru.