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Joint Entrance Examination
Mathematics
Definite Integration
Question 76
Question 76 of 246
Medium
MCQ
l
i
m
n
→
∞
3
n
4
+
(
2
+
1
n
)
2
+
(
2
+
2
n
)
2
+
…
+
(
3
−
1
n
)
2
\underset{n\to \infty}{lim}\frac{3}{n}{4+{(2+\frac{1}{n})}^{2}+{(2+\frac{2}{n})}^{2}+\ldots +{(3-\frac{1}{n})}^{2}}
n
→
∞
l
im
n
3
4
+
(
2
+
n
1
)
2
+
(
2
+
n
2
)
2
+
…
+
(
3
−
n
1
)
2
is equal to :
A
0
B
19
3
\frac{19}{3}
3
19
C
19
D
12
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Official solution
l
i
m
n
→
∞
∑
r
=
0
n
−
1
3
n
(
2
+
r
n
)
2
\underset{n\to \infty}{lim}\sum _{r=0}^{n-1}\frac{3}{n}{(2+\frac{r}{n})}^{2}
n
→
∞
l
im
∑
r
=
0
n
−
1
n
3
(
2
+
n
r
)
2
=
∫
1
0
3
(
2
+
x
)
2
d
x
={\int }_{1}^{0}3{(2+x)}^{2}\, dx
=
∫
1
0
3
(
2
+
x
)
2
d
x
=
3
.
(
2
+
x
)
3
3
∣
1
0
={3\, .\, \frac{{(2+x)}^{3}}{3}|}_{1}^{0}
=
3
.
3
(
2
+
x
)
3
∣
1
0
=
3
3
−
2
3
=
19
={3}^{3}-{2}^{3}=19
=
3
3
−
2
3
=
19
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