Let Sn=limn→∞(1n6+32n6+...+1n) and Tn=limn→∞(1n6+32n6+...+(n−1)5n6), then which of the following is/are true?
A
Sn→1+6
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
(Sn+Tn)<13
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(Sn+Tn)>13
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Tn→1−6
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct options are ASn→1+6
C(Sn+Tn)>13
DTn→1−6 Sn=limn→∞(1n6+32n6+243n6+...1n)Tn=limn→∞(1n6+32n6+243n6+....(n−1)5n6) Sn=limn→∞∑nr=1r5n6=∫10x5dx=x66|10=16 ⇒Sn=(16)+;Tn=(16)− But sincex5is concave upward the area included in Sn for any two consecutive values of r is more than area excluded in Tnfor same values of r ⇒Sn−16>16−Tn⇒Sn+Tn>13