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KEAM 2023 Math Solved Paper
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© examsnet.com
Question : 22
Total: 120
n
Σ
r
=
0
(
4
r
+
3
)
⋅
(
n
C
r
)
2
(
2
n
+
3
)
is
For any
n
≥
0
, the value of is
2
n
C
n
−
1
8
n
C
n
2
n
C
n
+
1
n
C
n
−
2
2
n
C
n
Validate
Solution:
n
∑
r
=
0
(
4
r
+
3
)
(
n
C
r
)
2
2
n
+
3
Substitute for
n
=
1
3
+
(
1
C
0
)
2
+
7
(
1
C
1
)
2
5
=
10
5
=
2
=
2
⋅
1
C
1
=
2
C
1
In the given choices option
E
matches with this
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