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Last updated at May 29, 2018 by Teachoo

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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)

Miscellaneous

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 You are here

Misc 24

Misc 25 Important

Misc 26

Misc 27

Misc 28 Important

Misc 29

Misc 30

Misc 31

Misc 32 Important

Chapter 9 Class 11 Sequences and Series

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.