JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023Among
(S1):limn→∞n21(2+4+6+⋯+2n)=1
(S2):limn→∞n161(115+215+315+⋯+n15)=161
Choose the correct answer:
- A.
Only (S1) is true.
- B.
Both (S1) and (S2) are true
(Correct Answer) - C.
Both (S1) and (S2) are false
Both (S1) and (S2) are true
Explanation
Solution for (S1):
S1:n→∞limn21(2+4+6+⋯+2n)
=n→∞limn22×2n(n+1)=n→∞lim(1+n1)=1
Solution for (S2):
and S2:n→∞limn161(115+215+315+⋯+n15)
=n→∞limn161r=1∑nr15=n→∞limn1r=1∑n(nr)15
=∫01x15dx=[16x16]01=161
Hence, both (S1) and (S2) are true.
Explanation
Solution for (S1):
S1:n→∞limn21(2+4+6+⋯+2n)
=n→∞limn22×2n(n+1)=n→∞lim(1+n1)=1
Solution for (S2):
and S2:n→∞limn161(115+215+315+⋯+n15)
=n→∞limn161r=1∑nr15=n→∞limn1r=1∑n(nr)15
=∫01x15dx=[16x16]01=161
Hence, both (S1) and (S2) are true.