IIT JEE-AIEEE 2012, Entrance Notifications, Study Tips, Engineering Preparation, Practice Test Papers, Experts Advice

Discussion Response Post to: Series

Forum Index -> Algebra -> View Full Question

Author

Message

Priyesh (1601)

Blazing goIITian

total posts: 1042
Offline

Hi A.Vignesh

the nth term of the series will be (1/2n-1)*(1/2n+1)

= 1/2[(1/2n-1) - (1/2n+1)]

now we write the first three terms and check if they cancel or not

taking 1/2 common we can write the terms as 1 - 1/3 , 1/3 - 1/5 , 1/5 - 1/7

therefore on adding we see that (-1/3 1/3),(-1/5 1/5).. will get cancelled leaving behind only 1 & -1/2n+1 (as they do not get cancelled)

therefore sum is 1/2[1 - 1/2n+1)

=1/2(2n/2n+1) = n/(2n+1)

therefore the sum of the series is n/2n+1

"Imagination is more important than knowledge."

Like 0 people liked this

Free Sign Up!

@ INR 1,995/-