questionr
Early access
Exams
Joint Entrance Examination
Mathematics
Indefinite Integrals
Question 205
Question 205 of 416
Medium
MCQ
The value of
lim
n
→
∞
1
n
∑
r
=
0
2
n
−
1
n
2
n
2
+
4
r
2
\displaystyle\lim_{n\to\infty}\frac1n\sum_{r=0}^{2n-1}\frac{n^2}{n^2+4r^2}
n
→
∞
lim
n
1
r
=
0
∑
2
n
−
1
n
2
+
4
r
2
n
2
is
A
1
2
tan
−
1
(
2
)
\dfrac12\tan^{-1}(2)
2
1
tan
−
1
(
2
)
B
1
2
tan
−
1
(
4
)
\dfrac12\tan^{-1}(4)
2
1
tan
−
1
(
4
)
C
tan
−
1
(
4
)
\tan^{-1}(4)
tan
−
1
(
4
)
D
1
4
tan
−
1
(
4
)
\dfrac14\tan^{-1}(4)
4
1
tan
−
1
(
4
)
Check answer
Press A-D to select · Enter to check
Solutions · 1
Post your solution
Official solution
\[\lim_{n\to\infty}\frac1n\sum_{r=0}^{2n-1}\frac{1}{1+4\left(\frac rn\right)^2}=\int_0^2\frac{dx}{1+4x^2}=\frac12\left[\tan^{-1}2x\right]_0^2=\frac12\tan^{-1}4\]
Answer the question to unlock
Previous
Next