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July 30, 2014

July 30, 2014

Posted by **louise** on Sunday, July 31, 2011 at 2:49am.

- algebra II -
**MathMate**, Sunday, July 31, 2011 at 4:06pm(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

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