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UPSC NDA 2 2017 Mathematics Solved Paper
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© examsnet.com
Question : 25
Total: 120
The value of
1
log
3
e
+
1
log
3
e
2
+
1
log
3
e
4
+
.
.
.
.
up to infinite terms is
[NDA II 2017]
log
e
9
0
1
log
e
3
Validate
Solution:
log
e
+
1
2
log
e
3
+
1
4
log
e
3
+
.
.
.
.
=
(
log
e
3
)
[
1
+
1
2
+
1
4
+
.
.
.
∞
]
=
(
log
e
3
)
1
1
−
1
2
=
2
log
e
3
=
log
e
9
© examsnet.com
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