+0

# (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.)

0
170
6
+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
+104414
-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
+85
+1

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

alskdj  Mar 14, 2019
#3
+103917
+2

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

Mar 14, 2019
edited by CPhill  Mar 14, 2019
#4
+104414
+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
+85
+1

Thanks, everyone!

Mar 15, 2019