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(a) For what positive integers $n$ does $\left(x^2+\frac{1}{x}\right)^n$ have a nonzero constant term?

(b) For the values of $n$ that you found in part (a), what is that constant term? (You can leave your answer in the form of a combination.)

 

PLease explain very well

 May 7, 2015

Best Answer 

 #2
avatar+129839 
+14

Amended answer...sorry, Mellie....I realized the answer I gave before was incorrect....

(a) It appears that a constant term will occur when n is a multiple of 3

For instance  

When n = 3, the constant term is 3  =  C(n, [2/3]n) = C(3,2)

When n = 6,  the constant term is 15  = C(6,4)  

When n = 9, the constant term is 84  = C(9,6)

 

(b) So  ...it appears that the constant term is produced by C(n,[2/3]n)    

 

This happens because at C(n, [2/3]n ).......x^1  is being raised to the [n - (2/3)n]  = (1/3)n power...and...

x^-1 is being raised to the (2/3)n power....

 

So   (x^2)^[(1/3)n]  * (x^-1)^[(2/3)n]   =   (x)^(2/3)n  * (x)^(-2/3)n  = 1

 

And  C(n, [2/3]n) * (1)   =  C(n, [2/3]n )   = a constant

 

 

  

 May 7, 2015
 #2
avatar+129839 
+14
Best Answer

Amended answer...sorry, Mellie....I realized the answer I gave before was incorrect....

(a) It appears that a constant term will occur when n is a multiple of 3

For instance  

When n = 3, the constant term is 3  =  C(n, [2/3]n) = C(3,2)

When n = 6,  the constant term is 15  = C(6,4)  

When n = 9, the constant term is 84  = C(9,6)

 

(b) So  ...it appears that the constant term is produced by C(n,[2/3]n)    

 

This happens because at C(n, [2/3]n ).......x^1  is being raised to the [n - (2/3)n]  = (1/3)n power...and...

x^-1 is being raised to the (2/3)n power....

 

So   (x^2)^[(1/3)n]  * (x^-1)^[(2/3)n]   =   (x)^(2/3)n  * (x)^(-2/3)n  = 1

 

And  C(n, [2/3]n) * (1)   =  C(n, [2/3]n )   = a constant

 

 

  

CPhill May 7, 2015

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