wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

Let Sn=nk=1nn2+kn+k2 and Tn=n1k=0nn2+kn+k2 for n=1,2,3, . Then,

A
Sn<π33
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Sn>π33
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Tn<π33
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Tn>π33
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct options are
A Sn<π33
D Tn>π33
Sn<limnSn=limnnk=11n11+k/n+(k/n)2
=10dx1+x+x2=π33
Now, Tn>π33 as hn1k=0f(kh)>10f(x)dx>hnk=1f(kh)
Ans: A,D

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Motion Under Gravity
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon