Solving the Binomial Theorem: (2x^2 - 1/x)^10

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Homework Help Overview

The discussion revolves around the application of the Binomial Theorem to the expression (2x^2 - 1/x)^10. Participants are exploring the general term, specific terms in the expansion, and the existence of certain terms such as x^5 and constant terms.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the general term and specific terms in the binomial expansion, questioning the existence of an x^5 term and a constant term.
  • Some participants clarify the process of identifying the 7th term and the nature of coefficients in the context of the binomial expansion.
  • Others suggest reconsidering the definitions of constant terms and the implications of powers of x in the expansion.

Discussion Status

Participants are actively engaging with the original poster's attempts, providing clarifications and corrections regarding the identification of terms and coefficients. There is a mix of interpretations regarding the existence of certain terms, and some productive guidance has been offered.

Contextual Notes

There is a mention of a potential misunderstanding regarding the definition of constant terms and the implications of fractional powers in the context of the binomial expansion. Additionally, a second, unrelated question about developing a formula is introduced, which shifts the focus slightly.

masterofthewave124
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1. For each of the following, simplify so that the variable term is raised is to a single power:
(a) State the General Term tr in the Binomial expansion of (2x^2 - 1/x)^10
(b) Find the 7th term in the expansion
(c) Is there an x^5 term? Find its coefficient.
(d) Is there a constant term [independent of x] ? Find it if there is?

this is what i have so far...

a) tr = C(10,r) (2x^2)^(10-r)*(-1/x)^r
= C(10,r) 2^(10-r) * x^(20-r)* (-1)^r * x^-r
= 2^(10-r) * (-1)^r * C(10,r) * x^(20-3r)

b) Since it is t7, r = 6

so t7 = 2^4 * (-1)^6 * C(10,6) * x^2
= 3360x^2

c) Let 20-3r = 5, so r = 5.

So t5 = 2^5 * (-1)^5 * C(10,5) * x^2
and the coefficient is -8064

d) Let 20-3r = 0
r = 20/3 so there is no constant term?


a somewhat unrelated topic, but another quesiton...

2. http://img161.imageshack.us/img161/4540/explicit9uq.jpg

how can i utilize the hints to develop a formula?
 
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b) So, you can find the 'general' term, but not the 7th? r goes from 0 to 10 in a) so the 7th term is when r=6. c) that is not a coefficient; the coefficient is the number that multiplies the x^5 term (see d))

d) if you can do c) you can do d) unless you just put r=5 because you saw x^5 and thought that automatically meant r=5. constant means x^0.
 
thanks matt, i edited my post around the same time you posted! can you check what i have now? and any help for the second question?
 
I would prefer to use the statement that (x^2)^r(1/x)^(10-r) always gives an odd power of x, hence there is no x^0 term. I dislike intensely unnecessary proofs that rely on showing something supposed to be an integer isn't by using fractions. Call me an old fuss pot if you must. But it is completely unnecessary to do this as a 'contradiction', as are many contradiction proofs.
 
Last edited:
For the second question, find the sum of the first n even squares (this should be easy), and then subtract this off the sum of the first 2n squares.
 
thanks for the information! more power
 

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