Asked by louise
find the 7th term of (x^3/2 y+b^2)^14
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MathMate
(x^3/2 y+b^2)^14
The coefficient of the ith term (i starts from zero) of a binomial expansion raised to the nth power is given by:
n!/[(n-i)!i!]
So the ith term (i=0,1,2,...) is given by:
n!/[n-i]!i!] * (x^(3/2)y)^(n-i)*b^(2i)
For n=14, and i=7 (the middle term):
T7=3432(x^(3/2)y)^7*b^14
=3432x^(21/2)y^7b^14
The coefficient of the ith term (i starts from zero) of a binomial expansion raised to the nth power is given by:
n!/[(n-i)!i!]
So the ith term (i=0,1,2,...) is given by:
n!/[n-i]!i!] * (x^(3/2)y)^(n-i)*b^(2i)
For n=14, and i=7 (the middle term):
T7=3432(x^(3/2)y)^7*b^14
=3432x^(21/2)y^7b^14
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