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Misc 25 - Find sum of 13/1 + 13 + 23/1 + 3 - Chapter 9 Class 11 - Finding sum from nth number

Misc 25 - Chapter 9 Class 11 Sequences and Series - Part 2
Misc 25 - Chapter 9 Class 11 Sequences and Series - Part 3 Misc 25 - Chapter 9 Class 11 Sequences and Series - Part 4 Misc 25 - Chapter 9 Class 11 Sequences and Series - Part 5

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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]

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.