Find 6th term in the expansion of (x^3 - 1/x^2)
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Answer:
The given term is (x
3
−
x
2
1
)
10
.
The general term of expansion of (a+b)
n
is,
T
r+1
=
n
C
r
a
n−r
b
r
The sixth term of the expansion of (x
3
−
x
2
1
)
10
is,
T
5+1
=
10
C
5
(x
3
)
10−5
(−
x
2
1
)
5
=−
5!(10−5)!
10!
(x
3
)
5
(
(x
2
)
5
1
)
=−
5!5!
10!
(x
15
)(
x
10
1
)
=−252x
15−10
=−252x
5
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