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CGPET 2019 Solved Paper
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Section:
Mathematics
© examsiri.com
Question :
135 of 150
Marks:
+1
,
-0
The value of
2
1
.
1
3
+
3
2
.
1
9
+
4
3
.
1
27
+
5
4
.
1
81
+
.
.
.
.
.
.
∞
is
1
2
−
log
e
2
3
−
log
e
2
3
1
2
+
log
e
2
3
log
e
2
3
Validate
Solution:
Let
S
=
2
1
.
1
3
+
3
2
.
1
9
+
4
3
.
1
27
+
5
4
.
1
81
+
.
.
.
.
.
.
+
n
+
1
n
.
1
3
n
+
.
.
.
.
.
.
∞
Where,
T
n
=
n
+
1
n
.
1
3
n
=
(
1
+
1
n
)
1
3
n
=
1
3
n
+
1
n
.
3
n
⇒
S
=
∑
T
n
=
∑
1
3
n
+
∑
1
n
.
3
n
=
1
3
1
−
1
3
+
{
−
log
e
(
1
−
1
3
)
}
=
1
2
−
log
e
(
2
3
)
© examsiri.com
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