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Test Index
TS EAMCET 4-Aug-2021 Shift 1 Question Paper
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Section:
Mathematics
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© examsnet.com
Question : 3 of 160
Marks:
+1
,
-0
For all
n
∈
N
,
1
2.5
+
1
5.8
+
.
.
.
+
1
(
3
n
−
1
)
(
3
n
+
2
)
=
n
6
n
+
4
n
2
6
n
+
4
1
2
.
n
2
6
n
+
4
n
6
n
2
+
4
Validate
Solution:
👈: Video Solution
1
2
⋅
5
=
1
3
(
1
2
−
1
5
)
,
1
5
⋅
8
=
1
3
(
1
5
−
1
8
)
,
.
.
.
,
1
(
3
n
−
1
)
(
3
n
+
2
)
=
1
3
[
1
3
n
−
1
−
1
3
n
+
2
]
∴
1
2
⋅
5
+
1
5
⋅
8
+
.
.
.
+
1
(
3
n
−
1
)
(
3
n
+
2
)
=
1
3
(
1
2
−
1
5
)
+
1
3
(
1
5
−
1
8
)
+
.
.
.
+
1
3
(
1
3
n
−
1
−
1
3
n
+
2
)
=
1
3
[
1
2
−
1
5
+
1
5
−
1
8
+
.
.
.
+
1
3
n
−
1
−
1
3
n
+
2
]
=
1
3
(
1
2
−
1
3
n
+
2
)
=
1
3
⋅
3
n
+
2
−
2
2
(
3
n
+
2
)
=
n
6
n
+
4
© examsnet.com
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