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JEE Main Math Class 11 Binomial Theorem Part 2 Questions
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© examsnet.com
Question : 7
Total: 115
Let
α
=
n
∑
r
=
0
(
4
r
2
+
2
r
+
1
)
n
C
r
and
β
=
(
n
∑
r
=
0
n
C
r
r
+
1
)
+
1
n
+
1
. If
140
<
2
α
β
<
281
, then the value of
n
is ______.
[8 Apr 2024 Shift 1]
Your Answer:
Validate
Solution:
👈: Video Solution
α
=
n
∑
r
=
0
(
4
r
2
+
2
r
+
1
)
⋅
n
C
r
α
=
4
∑
r
=
0
n
r
2
⋅
n
r
⋅
n
−
1
C
r
−
1
+
2
∑
r
=
0
n
r
⋅
n
r
⋅
n
−
1
C
r
−
+
n
∑
r
=
0
n
C
r
,
+
4
n
n
−
1
∑
r
=
0
n
C
r
−
1
+
2
n
n
−
1
∑
r
=
0
n
C
r
−
1
+
n
∑
r
=
0
n
C
r
α
=
4
n
(
n
∣
−
1
)
⋅
2
n
−
2
+
4
n
⋅
2
n
−
1
+
2
n
⋅
2
n
−
1
+
2
n
α
=
2
n
−
2
[
4
n
(
n
−
1
)
+
8
n
+
4
n
+
4
]
α
=
2
n
−
2
[
4
n
2
+
8
n
+
4
]
=
2
n
+
1
n
+
1
2
α
β
=
2
n
+
1
(
n
+
1
)
2
2
n
+
1
⋅
(
n
+
1
)
=
(
n
+
1
)
3
140
<
(
n
+
1
)
3
<
281
n
=
4
⇒
(
n
+
1
)
3
=
125
n
=
5
⇒
(
n
+
1
)
3
=
216
n
=
6
⇒
(
n
+
1
)
3
=
343
∴
n
=
5
© examsnet.com
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