# Misc 25 - Chapter 9 Class 11 Sequences and Series

Last updated at March 9, 2017 by Teachoo

Last updated at March 9, 2017 by Teachoo

Transcript

Misc , 25 Find the sum of the following series up to n terms: 13/1 + (13+23)/(1+3) + (13+23+33)/(1+3+5) +………. 13/1 + (13 + 23)/(1 + 3) + (13 + 23 + 33)/(1 + 3 + 5) +………. nth term of series is an = (13 + 23 + 33 + …. + n3)/(1 + 3 + 5 + … + 𝑛 𝑡𝑒𝑟𝑚𝑠) We solve denominator & numerator separately 13 + 23 + 33 + … + n3 = ((𝑛(𝑛+1))/2)^2 Also, 1 + 3 + 5 + …. + n terms This is A.P whose first term is a = 1 & common difference = d = 3 – 1 = 2 Now, sum of n terms of AP is Sn = n/2 [ 2a +(n – 1)d] Sn = n/2 [ 2(1) + (n-1)2] Sn = n/2 [ 2 + 2n - 2] Sn = 𝑛/2[2n] Sn = n2 Now, an = (13 + 23 + 23 + …. + n3)/(1 + 3 + 5 + … +𝑛 𝑡𝑒𝑟𝑚𝑠) Putting values from (1) & (2) an = (𝑛(𝑛 + 1)/2)^2/𝑛2 = n2(n+1)2/4n2 = (n+1)2/4 = ((n2+1+2𝑛))/4 = (𝑛2 + 1 + 2𝑛)/4 Now, finding sum of n terms of series = 1/4 ((((𝑛)(𝑛 +1)(2𝑛 +1))/6)" + " 2 (𝑛(𝑛 + 1))/2 "+ n" ) = 1/4 ((𝑛(𝑛+1)(2𝑛+1))/6 " + n(n +1) + n" ) = 1/4 n(( (𝑛+1)(2𝑛+1))/6 " + (n +1) + 1" ) = 1/4 n(( (𝑛+1)(2𝑛+1)+6(𝑛+1)+6(1))/6 " " ) = 1/4 n(( (𝑛+1)(2𝑛+1)+6(𝑛+1)+6(1))/6 " " ) = (1 × 𝑛)/(4 × 6) [(n + 1)(2n + 1) + 6(n + 1) + 6] = 𝑛/24 [n(2n + 1) + 1(2n + 1) + 6n + 6 + 6] = 𝑛/24 [2n2 + n + 2n + 1 + 6n + 6 + 6] = 𝑛/24 [2n2 + n + 2n + 6n + 1 + 6 + 6] = 𝑛/24 [2n2 + 9n + 13] Thus , the required sum is 𝑛/24 [2n2 + 9n + 13]

Misc 1

Misc 2

Misc 3 Important

Misc 4

Misc 5

Misc 6

Misc 7 Important

Misc 8

Misc 9

Misc 10

Misc 11

Misc 12

Misc 13

Misc 14

Misc 15

Misc 16 Important

Misc 17

Misc 18

Misc 19 Important

Misc 20

Misc 21

Misc 22

Misc 23

Misc 24

Misc 25 Important You are here

Misc 26

Misc 27

Misc 28 Important

Misc 29

Misc 30

Misc 31

Misc 32 Important

Chapter 9 Class 11 Sequences and Series

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.