find the co fficient of x5 in (x+3)^8
Answers
Answered by
2
Step-by-step explanation:
It is known that (r+1)th term (T
r+1
) in the binomial expansion of
(a+b)
n
is given by T
r+1
=
n
C
r
a
a−r
b
r
Assuming that x
5
occurs in the (r+1)th term of expansion (x+3)
8
we obtain
T
r+1
=
8
C
r
(x)
8−r
(3)
r
Comparing the indices of x in x
5
and in T
r+1
, we obtain r=3
Thus, the coefficient of x
5
is
8
C
3
(3)
3
=
3!5!
8!
x3
3
=
3.2.5!
8.7.6.5!
.3
3
=1512
Answered by
1
answer
It is known that (r + 1) th term (Tr + 1) in the binomial expansion of (a + b)n is given by Tr + 1 = nCr aa - rbr Assuming that x5 occurs in the (r + 1)
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