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Question
Mathematics
The coefficient of xn in the polynomial (x+nC0)(x+3. nC1)(x+5. nC2) .... (x+(2n+1)nCn) is
Q. The coefficient of
x
n
in the polynomial
(
x
+
n
C
0
)
(
x
+
3
.
n
C
1
)
(
x
+
5
.
n
C
2
)
....
(
x
+
(
2
n
+
1
)
n
C
n
)
is
2327
198
Binomial Theorem
Report Error
A
n
.
2
n
17%
B
n
.
2
n
+
1
28%
C
(
n
+
1
)
.
2
n
39%
D
(
n
+
1
)
.
2
n
+
1
17%
Solution:
(
x
+
n
C
0
)
(
x
+
3
.
n
C
1
)
(
x
+
5
.
n
C
2
)
....
(
x
+
(
2
n
+
1
)
.
n
C
n
)
=
x
n
+
1
+
x
n
{
n
C
0
+
3
.
n
C
1
+
5
.
n
C
2
+
.....
+
(
2
n
+
1
)
.
n
C
n
}
+
.....
Coeff. of
x
n
=
n
C
0
+
3
.
n
C
1
+
5
.
n
C
2
+
.....
+
(
2
n
+
1
)
.
n
C
n
=
1
+
(
n
C
1
+
2
.
n
C
1
)
+
(
n
C
2
+
4
.
n
C
2
)
+
....
+
(
n
C
n
+
2
n
.
n
C
n
)
=
(
1
+
n
C
1
+
.....
+
n
C
n
)
+
2
(
n
C
1
+
2
n
C
2
+
....
+
n
.
n
C
n
)
=
2
n
+
2
[
n
+
2.
2
!
n
(
n
−
1
)
+
3.
3
!
n
(
n
−
1
)
(
n
−
2
)
+
...
+
n
.1
]
=
2
n
+
2
n
[
1
+
n
−
1
C
1
+
n
−
1
C
2
+
.....
+
n
−
1
C
n
−
1
]
=
2
n
+
2
n
.
2
n
−
1
=
2
n
(
1
+
n
)
=
(
n
+
1
)
.
2
n