Determine the term containing (Binomial Theorom)

[zaddyzad]

Determine the term containing... (Binomial Theorom)

Homework Statement

Determine the term containing x^11 in the expansion (X^2 - 1/x)^10

Homework Equations

T(k+1) = nC(k)*A^(n-k)*b^K

The Attempt at a Solution

x^11 = 10Ck * (x^2)^(10-k)*(-X^-1)^k (taking out the (-) from the x^-1 b/c it don't matter)

x^11 = 10ck *X^(20-2k)(X^-k)
x^11/(X^20-3k) = 10ck
x^3k-9 = 10ck

Im stuck here.

 

[haruspex]

Start by finding the coefficient of x¹. That should give you a hint.

 

[zaddyzad]

Its just 1. How does that help me ?

 

[Dick]

Its just 1. How does that help me ?

I don't think it is 1. $(x^2-\frac{1}{10})^{10}$ contains only even powers of x, doesn't it? Forget the binomial theorem and just imagine what would happen if you start expanding it by hand.

 

[zaddyzad]

I don't think it is 1. $(x^2-\frac{1}{x})^{10}$ contains only even powers of x, doesn't it? Forget the binomial theorem and just imagine what would happen if you start expanding it by hand.

Mistake in my question.

 

[Dick]

Mistake in my question.

Yeah. So you want $(x^2-\frac{1}{x})^{10}$. Factor out the x^2. So now you've got $x^{20} (1-\frac{1}{x^3})^{10}$. What power do you want in the second factor?

 

[zaddyzad]

Yeah. So you want $(x^2-\frac{1}{x})^{10}$. Factor out the x^2. So now you've got $x^{20} (1-\frac{1}{x^3})^{10}$. What power do you want in the second factor?

Woah how did you get that? And I am looking for the term with X^11.

 

[Dick]

Woah how did you get that? And I am looking for the term with X^11.

I used algebra. And to go back to your first post, which was pretty hard to read especially with the wrong problem statement, you want 20-3k=11. What's k? And the sign does matter.

 

[zaddyzad]

Ok, so I'm looking for the term containing X^11. I tried using the formula T(k+1) = nC(k)*A^(n-k)*b^K. K is the variable of the formula.

 

[zaddyzad]

n is 10.

 

[zaddyzad]

I know the answer is 4 b/c it would be. 10c3(x^2)^7(-1/x)^3. However I don't care what the constant term is.

 

[zaddyzad]

K is 3 making the term 4 according to the formula T(k+1) = nC(k)*A^(n-k)*b^K.

 

[Dick]

Ok, so I'm looking for the term containing X^11. I tried using the formula T(k+1) = nC(k)*A^(n-k)*b^K. K is the variable of the formula.

You were almost there is the first post. You want the exponent in (x^2)^(10-k)*(x^(-k)) to be 11. What's k?

 

[Dick]

K is 3 making the term 4 according to the formula T(k+1) = nC(k)*A^(n-k)*b^K.

Yes, k=3. Now think a little about the sign of the coefficient.

 

[zaddyzad]

Yes its negative, however can't I just dismiss the coefficient since I'm only looking for its variable ?

 

[haruspex]

Woah how did you get that? And I am looking for the term with X^11.

Dick took a factor of x² out of each of the 10 brackets. That turned each bracket into 1-1/x³, and combining the 10 factors outside produced x²⁰.
Now imagine expanding the (1-1/x³)¹⁰. Suppose some term of that has the x power as 1/x^k. You want the one which, when multiplied by x²⁰, produces x¹¹.

 

[Dick]

Yes its negative, however can't I just dismiss the coefficient since I'm only looking for its variable ?

You threw away the sign and got 10C3 once you figured out k=3. But you shouldn't have thrown away the sign. It might be -10C3.

 

[zaddyzad]

Dick took a factor of x² out of each of the 10 brackets. That turned each bracket into 1-1/x³, and combining the 10 factors outside produced x²⁰.
Now imagine expanding the (1-1/x³)¹⁰. Suppose some term of that has the x power as 1/x^k. You want the one which, when multiplied by x²⁰, produces x¹¹.

I like how this was solved with mental thinking. But how would you solve it using sheer algebra ?

 

[zaddyzad]

You know, what you guys showed / helped me with is enough to solve similar questions. Thank you. I didnt know you can take x^2 out of (x^2 - 1/x)^10

 

[Dick]

I like how this was solved with mental thinking. But how would you solve it using sheer algebra ?

You very nearly already did it in the first post. Add the insight that k must be 3, don't ignore the sign and try to write it out in a coherent way.

 

[zaddyzad]

You very nearly already did it in the first post. Add the insight that k must be 3, don't ignore the sign and try to write it out in a coherent way.

I can't have the insight that K must be 3 b/c that's what I was trying to solve for with algebra.

 

[Dick]

I can't have the insight that K must be 3 b/c that's what I was trying to solve for with algebra.

If x^(20-3k) is going to be x^11, k must be 3. Ok, forget 'insight'. You can work that out with algebra.