Finding sum from nth number

Chapter 8 Class 11 Sequences and Series
Concept wise

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Misc 26 Show that (1 22 + 2 32 + + n (n + 1)2)/(12 2 + 22 3 + + n2 (n + 1)) = (3n + 5)/(3n + 1) Taking L.H.S (1 22 + 2 32 + + n (n + 1)2)/(12 2 + 22 3 + + n2 (n + 1)) We solve denominator & numerator separately Solving numerator Let numerator be S1 = 1 22 + 2 32 + + n (n + 1)2 nth term is n (n + 1)2 Let an = n(n + 1)2 = n(n2 + 1 + 2n) = n3 + n + 2n2 Now finding S1 = (( ( + 1))/2)^2 + 2(( ( +1)(2 +1))/6) + n(n+1)/2 = ( ( + 1))/2 (n(n+1)/2 " + " (2(2 +1))/3 " + 1" ) = ( ( + 1))/2 (( 3 ( +1) + 2 2(2 +1)+ 6)/6) = (n(n + 1))/(2 6)[3n(n + 1) + 4(2n + 1) + 6] = (n(n + 1))/12[3n2 + 3n + 8n + 4 + 6] = ( ( + 1))/12[3n2 + 11n + 10] = ( ( + 1))/12[3n2 + 5n + 6n + 10] = ( ( + 1))/12[n(3n + 5) + 2(3n + 5)] = ( ( + 1))/12[(n + 2)(3n + 5)] Thus, S1 = ( ( + 1))/12[(n + 2)(3n + 5)] Now solving denominator Let denominator be S2 = 12 2 + 22 3 + + n2 (n + 1) nth term is n2(n + 1) Let bn = n2(n + 1) bn = n3 + n2 Now, calculating S2 = (( ( + 1))/2)^2 + (( ( +1)(2 +1))/6) = ( ( + 1))/2 (n(n+1)/2 " + " ((2 +1))/3) = ( ( + 1))/2 (n(n+1)/2 " + " ((2 +1))/3) = ( ( + 1))/2 (( 3 ( +1) + 2 (2 +1))/6) = (n(n + 1))/(2 6) (3n(n + 1) + 2(2n + 1)) = (n(n + 1))/12 (3n2 + 3n + 2(2n + 1)) = (n(n + 1))/12 (3n2 + 3n + 4n + 2) = (n(n+1))/12 (3n2 + 7n +2) = (n(n+1))/12 (3n2 + 6n + n +2) = (n(n+1))/12 (3n(n + 2) + 1(n +2)) = (n(n+1)(n+2)(3n+1))/12 Thus, S2 = (n(n+1)(n+2)(3n+1))/12 Now, Taking L.H.S (1 22 + 2 32 + + n (n + 1)2)/(12 2 + 22 3 + + n2 (n + 1)) = 1/ 2 = ((n(n+1)(n+2)(3n+5))/12)/((n(n+1)(n+2)(3n+1))/12) = (n(n+1)(n+2)(3n+5))/12 12/(n(n+1)(n+2)(3n+1)) = (n(n+1)(n+2)(3n+5))/(n(n+1)(n+2)(3n+1)) = ((3n+5))/((3n+1)) = 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.