Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

jpknegtel

  • 4 years ago

Need to use the binomal therom to expand (1+2x)/(1-2x) Not sure where to start to get it into the right format.

  • This Question is Closed
  1. jpknegtel
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[(1+2x)/(1-2x)\]

  2. jpknegtel
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh. Need to go up too and including the term \[x ^{2}\]

  3. King
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so its x^2+2x+1/1-2x?

  4. jpknegtel
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Could you explain the steps? thank you very much for your timee!

  5. Zed
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Is that \[(1+2x) \times (1-2x)\] or \[\frac{(1+2x)}{(1-2x)}\]?

  6. jpknegtel
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I have been trying to do those Fractions but am unable to do it! Soo fustrating! Yes it is the the (1+2x) over (1-2x)

  7. Zed
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Okay give me a minute to work this out :)

  8. jpknegtel
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If it is any help the answer is \[(1+2x)(1+2x+4x ^{2})\]

  9. y2o2
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    (1+2x) over (1-2x) can never be equal to (1+2x)(1+2x+4x²) and you can assure that by substitution.

  10. Zed
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    \[(1+2x)(1-2x)^{-1}=(1+2x)(1^{-1}+-1*1^{-1-1}*-2x+\frac{-1(-1-1)}{2} (1)^{-1-2}(2x)^2+...)\]\[=(1+2x)(1^{-1}+2x+\frac{-1(-2)}{2} (1)^{-3}*4x^2+...)\]\[=(1+2x)(1+2x+\frac{2}{2} *1*4x^2+...)\]\[=(1+2x)(1+2x+4x^2+...)\] This is from this rule \[(a+b)^n=a^n+na^{n-1}b+\frac{n(n-1)}{2}a^{n-2}b^2+....\] Sorry it took so long :D

  11. dumbcow
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    either the answer is wrong or you forgot something when posting the problem i agree with y2o2

  12. Zed
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    http://mathworld.wolfram.com/NegativeBinomialSeries.html

  13. dumbcow
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @zed yes it becomes an infinite sum..is that the solution they are looking for? their answer stops after 4x^2

  14. Zed
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Yes they only had to do the terms until it reaches x^2 power

  15. dumbcow
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok thanks for clearing it up :)

  16. jpknegtel
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you very much guys! Clears things up!

  17. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy