Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Miscellaneous
Misc 2
Misc 3
Misc 4 Important
Misc 5
Misc 6
Misc 7 Important
Misc 8
Misc 9
Misc 10 Important
Misc 11 (i) Important
Misc 11 (ii)
Misc 12 Important
Misc 13
Misc 14 Important
Misc 15 Important
Misc 16
Misc 17 Important
Misc 18 Important
Question 1 Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 7 Deleted for CBSE Board 2024 Exams
Question 8 Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 10 Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams You are here
Question 12 Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 14 Deleted for CBSE Board 2024 Exams
Miscellaneous
Last updated at May 29, 2023 by Teachoo
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)