Solving Binomial Expansions: Coefficient of x^k in (2x-1/x)^2007

In summary, the conversation is about determining the coefficient of x^k in the expansion of (2x - 1/x)^2007. The person trying to solve the problem thought it would be the constant terms without any variables, but this doesn't work when factored out. The other person explains that x does not drop out of the expression and the power is x^(2007-2*k). They then clarify that (2x)^2007-k is equivalent to (2)^2007-k * x^2007-k and thank the other person for their help.
  • #1
brunie
62
0
Hi,
Im having some troubles with this binomial expansion...

Determine the coefficient of x^k, where k is any integer, in the expansion of (2x - 1/x)^2007.

I figured it would just be
C(2007,k) * (2x)^2007-k * (-1/x)^k
= C(2007,k) * (2)^2007-k * x^2007-k * (-1/x)^k

therefore the coefficient would only be the constant terms with no variables

but when i tried it on a smaller scale (ie small exponent), and factored it out, the equation i found doesn't work

i think it is due to the common variable x

anyone kno what to do?
 
Physics news on Phys.org
  • #2
x doesn't drop out of that expression. The power is x^(2007-2*k).
 
  • #3
so
(2x)^2007-k isn't equivalent to (2)^2007-k * x^2007-k ??
 
  • #4
Sure it is. But x^(2007-k)*(1/x)^k=x^(2007-2*k).
 
  • #5
ok yes that makes sense
thanks for ur help
 

1. What is a binomial expansion and how is it used?

A binomial expansion is a mathematical technique used to simplify and expand expressions with two terms, also known as binomials. It involves using the binomial theorem to find the coefficients of each term in the expanded form of the expression.

2. How do you find the coefficient of a specific term in a binomial expansion?

To find the coefficient of a specific term in a binomial expansion, you can use the formula (n choose k), where n is the power of the binomial and k is the exponent of the term you want to find the coefficient for. In this case, we would use (2007 choose k) to find the coefficient of x^k in the expression (2x-1/x)^2007.

3. Can you provide an example of solving for the coefficient of x^k in a binomial expansion?

For example, if we wanted to find the coefficient of x^5 in the expansion of (2x-1/x)^10, we would use the formula (10 choose 5) which equals 252. Therefore, the coefficient of x^5 in the expansion is 252.

4. What is the importance of finding the coefficient of a specific term in a binomial expansion?

Finding the coefficient of a specific term in a binomial expansion allows for the simplification and evaluation of complex expressions. It also allows for the prediction of patterns and relationships between different terms in the expansion.

5. Are there any shortcut methods for finding the coefficient of a term in a binomial expansion?

Yes, there are several shortcut methods such as using Pascal's triangle or the multinomial theorem. These methods are useful for quickly finding the coefficients of terms in binomial expansions with larger powers or exponents.

Similar threads

Replies
12
Views
827
  • Calculus and Beyond Homework Help
Replies
6
Views
888
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Precalculus Mathematics Homework Help
Replies
3
Views
141
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Precalculus Mathematics Homework Help
Replies
5
Views
179
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
851
Back
Top