Let Sn=∑nl=1(l4+l3n+l2n2+2n4n5)andTn=∑n−1l=0(l4+l3n+l2n2+2n4n5),(n=1,2,3,...)then
A
Tn>16760
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Tn<16760
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Sn>16760
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Sn<16760
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is CSn>16760 Replace ln→xand1n→dx, The functoin we get is f(x)=x4+x3+x2+2 f(x) is increasing in (0,1) Sn=∑nl=1l4n4.1n+l3n3.1n+l2n2.1n+2n
(Snis the sum of all the rectangles >∫10(x4+x3+x2+2)dx(∵Sn>area under curve)=16760 ⇒Sn>16760
Tn=∑n−1l=0(l4+l3n+l2n2+2n4n5)(∵Tn<area under curve) ⇒Tn<∫10f(x)dx ⇒Tn<16760