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Test Index
TS EAMCET 4-Aug-2021 Shift 2 Question Paper
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Section:
Mathematics
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Ā© examsnet.com
Question : 3 of 160
Marks:
+1
,
-0
For all
n
ā
N
,
2
2
n
+
1
+
3
2
n
+
1
is divisible by
7
5
11
8
Validate
Solution:
Let
P
(
n
)
=
2
2
n
+
1
+
3
2
n
+
1
,
n
ā
N
ā
ā
=
2
2
n
ā
2
+
3
2
n
ā
3
ā
ā
=
4
n
ā
2
+
9
n
ā
3
ā
ā
=
2
ā
(
5
ā
1
)
n
+
3
ā
(
10
ā
1
)
n
By using bionomial expansion
(
5
ā
1
)
n
=
ā
n
C
0
5
n
ā
ā
n
C
1
5
n
ā
1
+
ā
n
C
2
5
n
ā
2
+
ā
n
C
n
(
ā
1
)
n
(
10
ā
1
)
n
=
ā
n
C
0
10
n
ā
ā
n
C
1
10
n
ā
1
+
.
.
.
+
ā
n
C
n
(
ā
1
)
n
=
2
[
n
C
0
5
n
ā
ā
n
C
1
5
n
ā
1
+
ā
n
C
2
5
n
ā
2
ā
ā
n
C
3
5
n
ā
3
+
.
.
.
+
(
ā
1
)
n
]
+
3
ā
[
ā
n
C
0
10
n
ā
ā
n
C
1
10
n
ā
1
+
ā
n
C
2
10
n
ā
2
ā
ā
n
C
3
10
n
ā
3
+
.
.
.
+
(
ā
1
)
n
]
=
2
ā
ā
n
C
0
5
n
ā
2
ā
ā
n
C
1
5
n
ā
1
+
2
ā
ā
n
C
2
5
n
ā
2
ā
2
ā
ā
n
C
3
5
n
ā
3
.
.
.
2
ā
(
ā
1
)
n
+
3
ā
ā
n
C
0
10
n
ā
3
ā
ā
n
C
1
10
n
ā
1
+
3
ā
ā
n
C
2
10
2
ā
n
ā
3
ā
ā
n
C
3
10
n
ā
3
.
.
.
+
3
ā
(
ā
1
)
n
=
ā
n
C
0
(
2
ā
5
n
+
3
ā
10
n
)
ā
ā
n
C
1
(
2
ā
5
n
ā
1
+
3
ā
10
n
ā
1
)
+
ā
n
C
2
(
2
ā
5
n
ā
2
+
3
ā
10
n
ā
2
)
ā
ā
n
C
3
(
2
ā
5
n
ā
3
+
3
ā
10
n
ā
3
)
+
.
.
.
+
(
ā
1
)
n
5
=
5
[
n
C
0
(
2
ā
5
n
ā
1
+
3
ā
2
n
ā
5
n
ā
1
)
ā
ā
n
C
1
(
2
ā
5
n
ā
2
+
3
ā
2
n
ā
1
ā
5
n
ā
2
)
+
ā
n
C
2
(
2
ā
5
n
ā
3
+
3
ā
2
n
ā
2
ā
5
n
ā
3
)
P
(
n
)
=
5
K
=
ā
C
3
(
2
ā
5
n
ā
4
+
3
ā
2
n
ā
3
ā
5
n
ā
4
)
+
.
.
.
+
(
ā
1
)
n
P
(
n
)
=
5
K
where,
K
=
ā
n
C
0
(
2
ā
5
n
ā
1
+
3
ā
2
n
5
n
ā
1
)
ā
ā
n
C
1
(
2
ā
5
n
ā
2
+
3
ā
2
n
ā
1
ā
5
n
ā
2
)
+
.
.
.
+
(
ā
1
)
n
]
ā“
2
2
n
+
1
+
3
3
n
+
1
is divisible by
5
,
ā
n
ā
N
.
Ā© examsnet.com
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