1. Chapter 8 Class 11 Sequences and Series
  2. Serial order wise
  3. Miscellaneous
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Question 12 - Miscellaneous - Chapter 8 Class 11 Sequences and Series

Last updated at Dec. 16, 2024 by Teachoo

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Question 12 If S1, S2, S3 are the sum of first n natural numbers, their squares and their cubes, respectively, show that 9S22 = S3 (1 + 8S1) It is Given that S1 is the sum of n natural numbers i.e. S1 = 1 + 2 + 3 + + n S1 = (n(n+1))/2 S2 is the sum of square of n natural numbers i.e. S2 = 12 + 22 + 32 + n2 S2 = (n(n+1)(2n+1))/6 Also S3 is the sum of their cubes i.e. S3 = 13 + 23 + 33 + n3 S3 = (n(n+1)/2)^2 S3 = n2(n+1)2/4 We need to show that 9S22 = S3 (1 + 8S1) Taking R.H.S S3 (1 + 8S1) = n2(n+1)2/4 ("1 + 8" ((n(n+1))/2)) = n2(n+1)2/4 (1 + 4n(n + 1)) = n2(n+1)2/4 (1 + 4n2 + 4n) = n2(n+1)2/4 ((2n)2 + (1)2 + 2 2n 1) = n2(n+1)2/4 (2n+1)2 = (n(n+1) (2n+1))^2/4 Taking L.H.S 9S22 = 9 ((n(n+1) (2n+1))/6)^2 = 9 ((n(n+1) (2n+1))^2/6^2 ) = 9 (n(n+1) (2n+1))^2/36 = (n(n+1) (2n+1))^2/4 = R.H.S Hence L.H.S = R.H.S Hence proved

  1. Chapter 8 Class 11 Sequences and Series
  2. Serial order wise

    3. Miscellaneous

    Arithmetic Progression →
Arithmetic Progression →
Chapter 8 Class 11 Sequences and Series
Serial order wise

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo