Questions
The sum of the series up to n terms is
Remember concepts with our Masterclasses.
80k Users 60 mins Expert Faculty Ask Questions
a
n(n+1)9n2+23n+136
b
n(n−1)9n2+23n+126
c
(n+1)9n2+23n+136
d
n9n2+23n+136
Ready to Test Your Skills?
Check Your Performance Today with our Free Mock Tests used by Toppers!detailed solution
Correct option is A
Let, Sn=1⋅3⋅5+2⋅5⋅8+3⋅7⋅11+…+nth term ∴ Tn=n(2n+1)(3n+2)∴ Sn=ΣTn=Σn(2n+1)(3n+2)=Σn6n2+7n+2=Σ6n3+7n2+2n=6Σn3+7Σn2+2Σn=6n(n+1)22+7n(n+1)(2n+1)6+2n(n+1)2=n(n+1)26n(n+1)2+7(2n+1)3+2=n(n+1)218n2+n+28n+14+126=n(n+1)218n2+46n+266=n(n+1)2×29n2+23n+136=n(n+1)9n2+23n+136Similar Questions
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
Popular Books
HC Verma Part 1HC Verma Part 2RS AgarwalRD SharmaLakhmir SinghDK GoelSandeep GargTS GrewalNCERT solutions
Class 12 NCERT SolutionsClass 11 NCERT SolutionsClass 10 NCERT SolutionsClass 9 NCERT SolutionsClass 8 NCERT SolutionsClass 7 NCERT SolutionsClass 6 NCERT Solutions
