A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

TrojanPoem

  • one year ago

Prove that: 2 * nC0 + 2^2 * nC1 / 2 + 2^3 * nC2/3 + 2^4 * nC3/4 + ... + 2^(n+1) * nCn / n+1 = 3^(n+1) -1 / n+1

  • This Question is Closed
  1. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    use latex friend

  2. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @dan815 , Hmm ganshie ?

  3. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What is latex friend ?

  4. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    where you type \\ [ \] before and after

  5. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1435164810600:dw|

  6. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    wow thats quite a find

  7. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    expand \(\large \dfrac{(1+2)^n}{n}\) using binomial thm ?

  8. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    oh ya xD

  9. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\frac{ 2 C_{0}^{n} }{ 1 } + \frac{ 2^2 C_{1}^{n} }{ 2 } + \frac{ 2^3 C_{3}^{n} }{ 3 } + \frac{ 2^4 C_{4}^{n} }{ 4} + ... + \frac{ 2^{n+1} C_{n}^{n} }{ n+1 } = \frac{ 3^{n+1} -1 }{ n+1 } \]

  10. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1435164975210:dw|

  11. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1435165082146:dw|

  12. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    they seem so close

  13. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    for startes thats just a constant we can put into the equation so

  14. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1435165254017:dw|

  15. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Do you really have to use this E ? I hate it.

  16. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    it just means u increment k by 1 everytime and add the new expression, id have to copy paste so much if we dont use it

  17. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1435165420765:dw|

  18. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    thats all i means

  19. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    here is somethign useful

  20. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    i think i saw this one identity before equation a summation of the summation of choose for 1 less with respect to +1 choosing

  21. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    let me find it, its in an older question i asked

  22. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok , brb Until you find it.

  23. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ok here are some useful ones we can go over the proofs for them depending on which ones we use

  24. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok

  25. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1435165847184:dw|

  26. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1435165991834:dw|

  27. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    the 2nd one is the one we just worked out fromm ganeshies suggesntion

  28. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    cause you think about expanding this bracket 2^n = (1+1)^n, this will just be that summation right

  29. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    oh baby i got an idea!! we can do those geometric series tricks here!! multiply by k and subtract the 2 sequences it will simplify i think

  30. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    show me

  31. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1435166233288:dw|

  32. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    can u show that one first, that is almost the same

  33. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @ganeshie8

  34. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I think ganshie is busy

  35. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1435166298788:dw|

  36. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ehh im chucking this xD

  37. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    wait not yet it might bw a useful idea

  38. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    if we can break it down into summations we know like some cobinations of the 2^n summation it looks so close to the gemetric series tricks

  39. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how about using derivatives, the right hand side is the derivative of x^n when x=2

  40. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\large (1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k\] differentiate both sides and get \[\large n(1+x)^{n-1}=\sum_{k=0}^n \binom{n}{k}kx^{k-1}\] let \(x=1\)

  41. 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

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.