# Misc 24 - Chapter 9 Class 11 Sequences and Series (Term 1)

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

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

Miscellaneous

Misc 1

Misc 2

Misc 3 Important

Misc 4

Misc 5

Misc 6 Important

Misc 7 Important

Misc 8

Misc 9

Misc 10 Important

Misc 11

Misc 12

Misc 13

Misc 14 Important

Misc 15

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Misc 17

Misc 18

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Misc 20

Misc 21 (i) Important

Misc 21 (ii)

Misc 22 Important

Misc 23 Important

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Misc 27

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Misc 29 Important

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.