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avatar+86 

(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.)

 

I need an INCREDIBLY thorough explanation please!!!!!

 

THANKS FOR ALL HELP

 

Im so cunfused i deseratly need help

 Mar 13, 2019
edited by alskdj  Mar 13, 2019
 #1
avatar+110068 
-2

a)  n can be any multiple of 3.

You have a habit of telling people that their answers are wrong.   

So if you are going to say this then you might as well say it now before I explain my answer.

 Mar 13, 2019
 #2
avatar+86 
+1

sorry... can you please explain? i dont quite understand. i dont think you are wrong?

alskdj  Mar 14, 2019
 #3
avatar+111430 
+2

To see Melody's  answer

 

Note that......when n = 1

 

(x^2 + 1/x)^(3*1)  = (x^2 + 1/x)^3  will have the term C(3,2)(x^2)^1 * (1/x)^2  = 3x^2(1/x^2)

Constant term = 3

 

When n = 2.....

(x^2 + 1/x)^(3*2) = (x^2 + 1/x)^6  will have the term C(6, 4)(x^2)^2 * (1/x)^4 = 15 (x^4)(1/x^4) 

Constant term = 15

 

When n = 3....

(x^2 + 1/x)(3 * 3)   will have the term  C(9, 6) (x^2)^3 *(1/x)^6  = 84 (x^6)(1/x^6) 

Constant term = 84

 

So.....it appears the the pattern for the constant term will be  

 

C(3n, 2n) * (x^2)^n * (1/x)^(2n)      where n is a positive integer

 

 

 

cool cool cool

 Mar 14, 2019
edited by CPhill  Mar 14, 2019
 #4
avatar+110068 
+2

ok here is my logic.

 

\((x^2+\frac{1}{x})^n\)

 

let a be an integer such that    \(0\le a \le n\)

 

the a'th term will be

 

\(\begin{pmatrix}n\\a\end{pmatrix}[x^2]^a\;\left[ \frac{1}{x} \right]^{n-a}\\ \begin{pmatrix}n\\a\end{pmatrix}x^{2a}\;\times x ^{a-n}\\ \begin{pmatrix}n\\a\end{pmatrix}x^{3a-n}\\\ \)

 

this will be a constant term whenever

  \(3a-n=0\\n=3a\)

 

Since a is an integer, the expression will have a constant term if n is a multiple of 3

The value of the constant term is already included in my answer.

 Mar 14, 2019
 #6
avatar+86 
+1

Thanks, everyone! 

 Mar 15, 2019

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