Miscellaneous

Chapter 9 Class 11 Sequences and Series
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Misc 24 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

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.