1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Binomial Identity

  1. Apr 28, 2009 #1
    Prove this using this identity:
    [tex]
    k\binom{n}{k}=n\binom{n-1}{k-1}
    [/tex]
    [tex]
    \binom{n}{1}-2\binom{n}{2}+3\binom{n}{3}+....+(-1)n-1\binom{n}{n}
    [/tex]

    I was able to do this via differentiation, but not using this substitution. Any hints would be great.
     
    Last edited: Apr 28, 2009
  2. jcsd
  3. Apr 28, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi chaotixmonjuish! :smile:

    (use ^ and _ not sup and sub in latex :wink:)
    Well, the n is the same all the way through, that leaves a sum which should be easy. :wink:
     
  4. Apr 28, 2009 #3
    what do you mean?
     
  5. Apr 28, 2009 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Show us what you get when you substitute. :smile:
     
  6. Apr 28, 2009 #5
    Well this is something I sort of worked out:

    [tex]
    \binom{n}{1}-2\binom{n}{2}+....+(-1)^k\binom{n}{n}
    [/tex]
    Using the binomial theorem
    [tex]
    \sum_{k=0}^{n}\binom{n}{k}x^{k}=(1+x)^{n}
    [/tex]
    Are you saying I can just throw in the identity without doing anything or would I have to multiply across with a k so that
    [tex]
    \sum_{k=0}^{n}k\binom{n}{k}x^{k-1}=k(1+x)^{n}
    [/tex]
    so that
    [tex]
    \sum_{k=1}^{n-1}n\binom{n-1}{k-1}=k(x)^{k-1}
    [/tex]
    [tex]
    \sum_{l=0}^{n}n\binom{n-1}{l}x^{k-1}
    [/tex]

    I'm ignoring 1 because 1 to any power is 1
     
    Last edited: Apr 28, 2009
  7. Apr 28, 2009 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Why are you making this so complicated? :redface:

    Start with [tex]
    \sum_{l=0}^{n-1}n\binom{n-1}{l}
    [/tex] …

    what (in ordinary language) is (i'm leaving out the n :wink:) …

    [tex]
    \sum_{l=0}^{n-1}\binom{n-1}{l}
    [/tex] ? :smile:
     
  8. Apr 28, 2009 #7
    Well k-1=n, if that is what you are hinting at.
     
  9. Apr 29, 2009 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    No … what (in ordinary language) is [tex]\binom{n-1}{l}
    [/tex] (or n-1Cl) …

    it's the number of … ? :smile:
     
  10. Apr 29, 2009 #9
    Its sort of like saying its the number of ways you can select a leader (n), then how many different ways you cant form a team for leader [tex]\binom{n-1}{l}[/tex]
     
  11. Apr 29, 2009 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Yes …

    n-1Cl is the number of ways of choosing l things from n-1 …

    so (to get the ∑) what is the number of ways of choosing 1 thing or 2 things or … n-1 things from n-1? :smile:
     
  12. Apr 29, 2009 #11
    so would the final answer be

    [tex]
    n\sum_{k=1}^{n-1} \binom{n-1}{l}x^{l-1}
    [/tex]
     
    Last edited: Apr 29, 2009
  13. Apr 30, 2009 #12

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    i said you were making this complicated :rolleyes:

    wherever did x come from? :confused:

    (and you never answered:
    )
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Binomial Identity
  1. Binomial identity. (Replies: 3)

  2. Binomial identity (Replies: 1)

  3. Binomial identities (Replies: 2)

  4. A binomial identity (Replies: 8)

  5. Binomial Identity (Replies: 1)

Loading...