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Byju's Answer
Standard XII
Mathematics
Sum of n Terms
The sum of th...
Question
The sum of the series
1
1.2
−
1
2.3
+
1
3.4
−
.
.
∞
is
A
l
o
g
e
2
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B
2
l
o
g
e
2
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C
l
o
g
e
2
−
1
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D
l
o
g
e
4
e
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Solution
The correct option is
A
l
o
g
e
4
e
Let given series:
s
=
1
1.2
−
1
2.3
+
1
3.4
−
.
.
.
∞
let:
s
1
=
1
1.2
+
1
3.4
+
1
5.6
+
.
.
.
∞
t
n
=
1
(
2
n
−
1
)
(
2
n
)
t
n
=
1
2
n
−
1
−
1
2
n
∴
∑
s
n
=
∑
t
n
∑
s
n
=
∑
(
1
2
n
−
1
−
1
2
n
)
t
n
=
1
−
1
2
+
1
3
−
1
4
+
1
5
−
.
.
.
∞
t
n
=
l
o
g
e
2
......Eq:01
Solve:
s
2
=
1
2.3
+
1
4.5
+
1
6.7
+
.
.
.
∞
∴
∑
s
n
=
∑
t
′
n
∑
s
n
=
∑
(
1
2
n
−
1
2
n
+
1
)
t
′
n
=
1
(
2
n
)
(
2
n
+
1
)
t
′
n
=
1
2
n
−
1
2
n
+
1
t
′
n
=
(
1
2
−
1
3
)
+
(
1
3
−
1
4
)
+
.
.
.
∞
t
′
n
=
1
−
l
o
g
e
2
......Eq:02
t
′
n
=
−
(
−
1
2
+
1
3
−
1
4
+
.
.
.
∞
)
t
′
n
=
−
(
l
o
g
e
2
−
1
)
t
′
n
=
1
−
l
o
g
e
2
......Eq:02
Required summation of series (s)= Eq:01-Eq:02;
s
=
t
′
n
−
t
n
s
=
l
o
g
e
2
−
1
+
l
o
g
e
2
s
=
2
l
o
g
e
2
s
=
l
o
g
e
2
2
s
=
l
o
g
e
4
s
=
l
o
g
4
e
Option (D) is correct.
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