Determine the term containing (Binomial Theorom)

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The discussion focuses on determining the term containing x^11 in the expansion of (x^2 - 1/x)^10 using the Binomial Theorem. Participants clarify that the correct approach involves factoring out x^2, leading to the expression x^20(1 - 1/x^3)^10. They conclude that to find the term with x^11, k must equal 3, which is derived from the equation 20 - 3k = 11. There is a debate about whether to consider the sign of the coefficient in the final answer, with some emphasizing its importance. The conversation ultimately highlights the blend of algebraic manipulation and conceptual understanding needed to solve the problem effectively.
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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.
 
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Start by finding the coefficient of x1. That should give you a hint.
 


Its just 1. How does that help me ?
 


zaddyzad said:
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.
 


Dick said:
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.
 


zaddyzad said:
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?
 


Dick said:
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.
 


zaddyzad said:
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.
 


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.
 
  • #10


n is 10.
 
  • #11


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.
 
  • #12


K is 3 making the term 4 according to the formula T(k+1) = nC(k)*A^(n-k)*b^K.
 
  • #13


zaddyzad said:
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?
 
  • #14


zaddyzad said:
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.
 
  • #15


Yes its negative, however can't I just dismiss the coefficient since I'm only looking for its variable ?
 
  • #16


zaddyzad said:
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/xk. You want the one which, when multiplied by x20, produces x11.
 
  • #17


zaddyzad said:
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.
 
  • #18


haruspex said:
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/xk. 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 ?
 
  • #19


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
 
  • #20


zaddyzad said:
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.
 
  • #21


Dick said:
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.
 
  • #22


zaddyzad said:
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.
 
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