find the coefficient of x³ in the expansion of (2-3x)⁷
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Answer:
this is not exact answer dear
Step-by-step explanation:
Let a=((1+2x+3x
2
)
6
+(1−4x
2
)
6
)
∴ Coefficient of x
2
in the expansion of the product (2−x)
2
((1+2x+3x
2
)
6
+(1−4x
2
)
6
)
=2( Coefficient of x
2
in a)−1(Constant of expansion)
In the expansion of ((1+2x+3x
2
)
6
+(1−4x
2
)
6
),
Constant =1+1=2
Coefficient of x
2
= Coefficient of x
2
in (
6
C
0
(1+2x)
6
(3x
2
)
0
)+ Coefficient of x
2
in (
6
C
1
(1+2x)
5
3)−
6
C
1
(4)
=
6
C
2
4+6×3−24
=6=+18−24=54
Then, coefficient of x
2
in (2−x)
2
((1+2x+3x
2
)
6
+(1−4x
2
)
6
)
=2×54−1(2)=108−2=106
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