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Saturday, 31 October

JEE Main · Maths · Sequences and Series

Method of Differences for Telescoping Series

JEE Main asked Method of Differences for Telescoping Series 19 times in 19 of the 108 shifts Lumi analysed, in 2021, 2022, 2023, 2024, 2025 and 2026. It belongs to AP and GP, in Sequences and Series (Maths). The longest wait between two years asking it was 1 year. In 2021, 2022, 2023, 2024 and 2025 it came in more than one shift of the same year.

Updated 2026-10-24

Times asked
19
in 19 shifts
Years
6 of 8
First asked
2021
last in 2026
Pattern
Asked most years
across the analysed shifts

Questions on this idea a shift, by year

2017–2026. Asked in 2021, 2022, 2023, 2024, 2025 and 2026. Years differ in how many shifts Lumi analysed, so each year is scaled to one shift of 25 Maths questions.

00.150.30’170’200.10’210.18’220.28’230.19’240.25’250.11’26

Raw count: 2017: 0 in 1 shift · 2020: 0 in 12 shifts · 2021: 3 in 24 shifts · 2022: 5 in 24 shifts · 2023: 4 in 12 shifts · 2024: 4 in 18 shifts · 2025: 2 in 8 shifts · 2026: 1 in 9 shifts.

Shift by shift

Each square is one analysed shift, in date order. Open a filled one for that paper.

  • 20261 of 9
  • 20252 of 8
  • 20244 of 18
  • 20234 of 12
  • 20225 of 24
  • 20213 of 24
  • 20200 of 12
  • 20170 of 1

Asked in that shift More than once Not asked

How JEE Main framed it

Single correct is the usual form. Skill tested: calculation.

  • Sum of telescoping series
  • Sum of series with summation general term

Difficulty

Easy 3Medium 12Hard 4

4 of the 19 questions had a numerical answer.

Every question

All 19 questions

From the official JEE Main shift papers, newest first. Open one for the full question, the options and the official answer.

  1. 5 Apr 2026, Shift 1 · Q5∑n=110(528n(n+1)(n+2))\sum_{n=1}^{10}\left(\frac{528}{n(n+1)(n+2)}\right) is equal to:EasySingle correct
  2. 4 Apr 2025, Shift 2 · Q5If the sum of the first 20 terms of the series…MediumSingle correct
  3. 3 Apr 2025, Shift 1 · Q7The sum 1+3+11+25+45+71+…1 + 3 + 11 + 25 + 45 + 71 + \ldots upto 20 terms, is equal toMediumSingle correct
  4. 31 Jan 2024, Shift 1 · Q5The sum of the series 11−3⋅12+14+21−3⋅22+24+31−3⋅32+34+…\frac{1}{1-3\cdot1^2+1^4}+\frac{2}{1-3\cdot2^2+2^4}+\frac{3}{1-3\cdot3^2+3^4}+\ldots up to 10-terms isMediumSingle correct
  5. 30 Jan 2024, Shift 1 · Q24Let α=12+42+82+132+192+262+…\alpha=1^2+4^2+8^2+13^2+19^2+26^2+\ldots upto 10 terms and β=∑n=110n4\beta=\sum_{n=1}^{10}n^4. If 4α−β=55k+404\alpha-\beta=55k+40, then kk is equal…MediumNumerical value
  6. 9 Apr 2024, Shift 2 · Q24If…HardNumerical value
  7. 5 Apr 2024, Shift 1 · Q6If 11+2+12+3+…+199+100=m\frac{1}{\sqrt1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m and…EasySingle correct
  8. 12 Apr 2023, Shift 1 · Q8Let ⟨an⟩\langle a_n\rangle be a sequence such that a1+a2+…+an=n2+3n(n+1)(n+2)a_1+a_2+\ldots+a_n=\frac{n^2+3n}{(n+1)(n+2)}. If…HardSingle correct
  9. 10 Apr 2023, Shift 2 · Q8If Sn=4+11+21+34+50+…S_n = 4 + 11 + 21 + 34 + 50 + \ldots to nn terms, then 160(S29−S9)\frac{1}{60}(S_{29} - S_9) is equal toMediumSingle correct
  10. 8 Apr 2023, Shift 2 · Q8Let ana_n be the nthn^{th} term of the series 5+8+14+23+35+50+…5 + 8 + 14 + 23 + 35 + 50 + \ldots and Sn=∑k=1nakS_n = \sum_{k=1}^{n} a_k. Then S30−a40S_{30} - a_{40}…MediumSingle correct
  11. 6 Apr 2023, Shift 1 · Q7The sum of the first 20 terms of the series 5+11+19+29+41+…5+11+19+29+41+\ldots isMediumSingle correct
  12. 29 Jul 2022, Shift 2 · Q7∑r=120(r2+1)(r!)\sum_{r=1}^{20}\left(r^2+1\right)(r!) is equal toMediumSingle correct
  13. 29 Jul 2022, Shift 1 · Q4If 1(20−a)(40−a)+1(40−a)(60−a)+…+1(180−a)(200−a)=1256\frac{1}{(20 - a)(40 - a)} + \frac{1}{(40 - a)(60 - a)} + \ldots + \frac{1}{(180 - a)(200 - a)} = \frac{1}{256}, then the maximum…MediumSingle correct
  14. 27 Jul 2022, Shift 2 · Q24…HardNumerical value
  15. 26 Jul 2022, Shift 2 · Q24If ∑k=110kk4+k2+1=mn\sum_{k=1}^{10}\frac{k}{k^4+k^2+1}=\frac{m}{n}, where mm and nn are co-prime, then m+nm+n is equal to __________HardNumerical value
  16. 25 Jul 2022, Shift 2 · Q5The sum ∑n=1213(4n−1)(4n+3)\sum_{n=1}^{21}\frac{3}{(4n-1)(4n+3)} is equal toEasySingle correct
  17. 18 Mar 2021, Shift 1 · Q68132−1+152−1+172−1+…+1(201)2−1\frac{1}{3^2-1}+\frac{1}{5^2-1}+\frac{1}{7^2-1}+\ldots+\frac{1}{(201)^2-1} is equal to :MediumSingle correct
  18. 31 Aug 2021, Shift 1 · Q66The sum of 10 terms of the series 312×22+522×32+732×42+...\frac{3}{1^2\times2^2}+\frac{5}{2^2\times3^2}+\frac{7}{3^2\times4^2}+... is :MediumSingle correct
  19. 26 Aug 2021, Shift 2 · Q65lim⁡x→2(∑n=19xn(n+1)x2+2(2n+1)x+4)\lim_{x\to2}\left(\sum_{\mathrm{n}=1}^{9}\frac{x}{\mathrm{n}(\mathrm{n}+1)x^2+2(2\mathrm{n}+1)x+4}\right) is equal to :MediumSingle correct

Shifts: 18 Mar 2021, Shift 1, 26 Aug 2021, Shift 2, 31 Aug 2021, Shift 1, 25 Jul 2022, Shift 2, 26 Jul 2022, Shift 2, 27 Jul 2022, Shift 2, 29 Jul 2022, Shift 2, 29 Jul 2022, Shift 1, 6 Apr 2023, Shift 1, 8 Apr 2023, Shift 2, 10 Apr 2023, Shift 2, 12 Apr 2023, Shift 1, 30 Jan 2024, Shift 1, 31 Jan 2024, Shift 1, 5 Apr 2024, Shift 1, 9 Apr 2024, Shift 2, 3 Apr 2025, Shift 1, 4 Apr 2025, Shift 2, 5 Apr 2026, Shift 1.

Same topic

Other ideas in AP and GP

AP and GP has 23 ideas JEE Main has asked; 16 of them came up in more than one question. The 12 most asked are below.

Topic analysisChapter analysisStudy this chapterAll repeated ideas