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BITSAT 2017 Slot 2 Solved Paper
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© examsnet.com
Question : 106
Total: 150
Part IV
Mathematics
The coefficient of
x
5
in the expansion of
(
1
+
x
)
21
+
(
1
+
x
)
22
+
.
.
.
.
.
+
(
1
+
x
)
30
is
51
C
5
9
C
5
31
C
6
−
21
C
6
30
C
5
+
20
C
5
Validate
Solution:
(
1
+
x
)
21
+
(
1
+
x
)
22
+
.
.
.
.
.
+
(
1
+
x
)
30
=
(
1
+
x
)
21
[
1
+
(
1
+
x
)
1
+
.
.
.
.
+
(
1
+
x
)
9
]
=
(
1
+
x
)
21
[
(
1
+
x
)
10
−
1
(
1
+
x
)
−
1
]
=
1
x
[
(
1
+
x
)
31
−
(
1
+
x
)
21
]
∴
Coefficient of
x
5
in the given expression
=
Coefficient of
x
5
in
1
x
[
(
1
+
x
)
31
−
(
1
+
x
)
21
]
=
Coefficient of
x
6
in
[
(
1
+
x
)
31
−
(
1
+
x
)
21
]
=
31
C
6
−
21
C
6
© examsnet.com
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