Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Let s n =∑k=1 to n[n/(n 2 +kn+k 2 )] and T n =∑k=0 to (n-1) [n/(n 2 +nk+k 2 )] for n=1,2,3....... Then A S n ‹ pi/3√3 C T n B Sn> pi/3 √3 D T n>pi /3√3

Let sn=∑k=1 to n[n/(n2+kn+k2)] and Tn=∑k=0 to (n-1)  [n/(n2+nk+k2)] for n=1,2,3....... Then


A  Sn‹ pi/3√3                          C  Tn

B  Sn> pi/3√3                           D  Tn>pi/3√3

Grade:Upto college level

1 Answers

Badiuddin askIITians.ismu Expert
147 Points
12 years ago

Hi vanya

You can find Sn by summation of series using definite intigral as the limit of sum

sn=∑k=1 to n[n/(n2+kn+k2)]

sn=1/n∑k=1 to n [n2/(n2+kn+k2)]

sn=1/n∑k=1 to n [1/(1+k/n+(k/n)2)]

sn =∫1/(1+x+x2)dx   limit 0 to 1

   after solving

sn =pi/3√3


Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
 We are all IITians and here to help you in your IIT JEE preparation. All the best vanya
 
 

Regards,

Askiitians Experts

BAdiuddin


   


Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free