We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
135
6
avatar+85 

(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+102763 
-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+85 
+1

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

alskdj  Mar 14, 2019
 #3
avatar+102320 
+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+102763 
+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+85 
+1

Thanks, everyone! 

 Mar 15, 2019

4 Online Users