((3/x)+x2)8
a.) number of terms ? 9
b.) determine the middle term in the expansion, first in unsimplified terms and then in simplified ?
c.)will the expansion have a constant term ?Explain
Thanks in advance
+118723 Yes the number of terms is 9 
r=0,1,2,......9 The middle on is the 5th one which is r=4
((3/x)+x2)8
$$\mbox{The middle term is }\quad 8C4\times \left(\dfrac{3}{x}\right)^4\times(x^2)^4$$
You can simplify this yourself.
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$$\mbox{The general term is }\quad 8Cr\times \left(\dfrac{3}{x}\right)^r\times(x^2)^{8-r}$$
Where: $$0\le r\le 8\qquad \mox{ and }\qquad r \in Z$$
$$\mbox{This simplifies down to }\quad 8Cr\times 3^r\times x^{16-2r-r}=8Cr\times 3^r\times x^{16-3r}$$
Now if there is a constant term it will be when 16-3r=0 but the solution to this gives r as a non-integer.
So there must not be any constant term!
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+118723 Yes the number of terms is 9 r=0,1,2,......9 The middle on is the 5th one which is r=4
((3/x)+x2)8
$$\mbox{The middle term is }\quad 8C4\times \left(\dfrac{3}{x}\right)^4\times(x^2)^4$$
You can simplify this yourself.
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$$\mbox{The general term is }\quad 8Cr\times \left(\dfrac{3}{x}\right)^r\times(x^2)^{8-r}$$
Where: $$0\le r\le 8\qquad \mox{ and }\qquad r \in Z$$
$$\mbox{This simplifies down to }\quad 8Cr\times 3^r\times x^{16-2r-r}=8Cr\times 3^r\times x^{16-3r}$$
Now if there is a constant term it will be when 16-3r=0 but the solution to this gives r as a non-integer.
So there must not be any constant term!
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