the coefficient of x^12 in the expansion of (1 + 2x^2 - x^3)^8 is
Answers
Answer:
Correct option is
B
84
(1+r
2
−x
3
)
2
We need to find coefficient of x
6
First just let us simply the equation
f(x)=(1+x
2
(1−x))
8
now, we should right general term of a Binomial is
T
r
=
8
C
r
[1]
r
[x
2
(1−x)]
8−r
In this case it [1]
r
=1
∴T
r
=
8
C
μ
[x
2
(1−x)]
8−r
now r can take value from 0−8
∴ At r=1,T
1
=
8
C
1
[x
2
(1−x)]
9
=
8
C
1
x
14
(1−x)
7
It is clearly do not contain x
6
term as x
At r=2,T
2
=
8
C
2
[x
2
(1−x)]
6
no x
6
term
Similarly we will get x
6
at r=5,6
At r=5,T
5
=
8
C
5
[x
2
(1−x)]
3
=x
6
(1−x)
3
8
C
r
Since when x
6
will multiple of it will yield
Similarly in r=6,T
6
=
8
C
6
[x
2
(1−x)]
2
=x
2
(1−x)
2
When x
4
multiplies by x
2
it yield
8
C
6
x
6
∴ Their will be two term
8
C
5
x
6
+
8
C
6
x
6
It will at x
6
=
8
C
5
+
8
C
6
=84
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