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Misc 23 - Find sum of series: 3 + 7 + 13 + 21 + 31 + … - Miscellaneous

Misc 23 - Chapter 9 Class 11 Sequences and Series - Part 2
Misc 23 - Chapter 9 Class 11 Sequences and Series - Part 3
Misc 23 - Chapter 9 Class 11 Sequences and Series - Part 4

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Misc 23 Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + … It is not an AP or a GP Let Sn = 3 + 7 + 13 + 21 + 31 + …+ an–1 + an Sn = 0 + 3 + 7 + 13 + 21 + …. + an – 2 + an – 1 + an Subtracting (2) from (1) Sn – Sn = 3 – 0 + [(7 – 3) + (13 – 7) + (21 – 13) + … + (an-1 – an – 2) + (an – an – 1)] – an 0 = 3 + [4 + 6 + 8 + … an –1 ] – an an = 3 + [4 + 6 + 8 + … + an –1 ] Now 4 + 6 + 8 + … + an – 1 is an AP Whose first term a = 4 common difference d = 6 – 4 = 2 We know that, Sum of n terms of AP = 𝑛/2 (2a + (n – 1)d) Putting n = n – 1 , a = 4 , d = 2 [4 + 6 + 8 + … (n –1) terms] = ((n − 1)/2) [ 2a +(n – 1 – 1)d] = ((n − 1)/2) [ 2(4) +(n – 2)2 ] = ((n − 1)/2) [ 8 + 2n – 4 ] = ((n − 1)/2) [ 2n + 4 ] = (n−1)/2 × 2(n+2) = (n – 1) (n + 2) Thus, [4 + 6 + 8 + … upto (n –1) terms] = (n – 1) (n + 2) From (3) an = 3 + [4 + 6 + 8 + … + an –1 ] Putting values = 3 + (n – 1) (n + 2) = 3 + n(n + 2) – 1(n + 2) = 3 + n2 + 2n – n – 2 = n2 + n + 3 – 2 = n2 + n + 1 Now = (n(n + 1)(2n + 1))/6 + (n(n + 1))/2 + n = (n(n + 1)(2n + 1) + 3n(n + 1) + 6n)/6 = n(((n + 1)(2n + 1) + 3(n + 1) + 6)/6) = n ((2n2 + n + 2n + 1 + 3n + 3 + 6)/6) = n ((2n2+6n+10)/6) = n/6 × 2 (n2 + 3n +5) = n/3 (n2 + 3n +5) Thus, the required sum is n/3 (n2 + 3n +5)

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