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JEE Main Math Class 11 Binomial Theorem Part 2 Questions
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© examsnet.com
Question : 4
Total: 115
If the constant term in the expansion of
(
1
+
2
x
−
3
x
3
)
(
3
2
x
2
−
1
3
x
)
9
is
p
, then
108
p
is equal to
[5 Apr 2024 Shift 1]
Your Answer:
Validate
Solution:
👈: Video Solution
(
1
+
2
x
−
3
x
3
)
(
3
2
x
2
−
1
3
x
)
9
General term
m
(
3
2
x
2
−
1
3
x
)
9
=
9
C
r
⋅
3
9
−
2
r
2
9
−
r
(
−
1
)
r
⋅
x
18
−
3
r
Put
r
=
6
to get coeff. of
x
0
=
9
C
6
⋅
1
6
3
⋅
x
0
=
7
18
x
0
Put
r
=
7
to get coeff. of
x
−
3
=
9
C
r
⋅
3
−
5
2
2
(
−
1
)
7
⋅
x
−
3
=
−
9
C
7
⋅
1
3
5
⋅
2
2
⋅
x
−
3
=
−
1
27
x
−
3
(
1
+
2
x
−
3
x
3
)
(
7
18
x
0
−
1
27
x
−
3
)
7
18
+
3
27
=
7
18
+
1
9
=
7
+
2
18
=
9
18
=
1
2
∴
108
⋅
1
2
=
54
© examsnet.com
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