SOLUTION: Please find the middle term in the expansion of (x/3 + 9y)^10

Question 171588: Please find the middle term in the expansion of
(x/3 + 9y)^10
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
.
You will need to apply the "binomial theorem"
See this site for a review:
http://www.purplemath.com/modules/binomial.htm
.
The nth term is:
C(n,k)a^(n-k)b^k
where
C(n,k) = n!/(n-k)!k!
.
In this case,
k = 4 (desired term minus 1)
n = 10
.
therefore,
C(10,4)a^(10-4)b^4
.
C(n,k) = n!/(n-k)!k!
C(10,4) = 10!/(10-4)!4!
C(10,4) = 10!/6!4!
C(10,4) = (7*8*9*10)/(1*2*3*4) = 120960
.
So,
C(10,4)a^(10-4)b^4
120960a^6b^4
120960(x/3)^6(9y)^4
120960(x^6/729)(9y)^4
165.9259(x^6)(6561y^4)
1088640x^6y^4

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