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Binomial Identity

  • #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:

Answers and Replies

  • #2
tiny-tim
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Hi chaotixmonjuish! :smile:

(use ^ and _ not sup and sub in latex :wink:)
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.
Well, the n is the same all the way through, that leaves a sum which should be easy. :wink:
 
  • #3
what do you mean?
 
  • #4
tiny-tim
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  • #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:
  • #6
tiny-tim
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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:
 
  • #7
Well k-1=n, if that is what you are hinting at.
 
  • #8
tiny-tim
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No … what (in ordinary language) is [tex]\binom{n-1}{l}
[/tex] (or n-1Cl) …

it's the number of … ? :smile:
 
  • #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]
 
  • #10
tiny-tim
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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]
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:
 
  • #11
so would the final answer be

[tex]
n\sum_{k=1}^{n-1} \binom{n-1}{l}x^{l-1}
[/tex]
 
Last edited:
  • #12
tiny-tim
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so would the final answer be

[tex]
n\sum_{k=1}^{n-1} \binom{n-1}{l}x^{l-1}
[/tex]
i said you were making this complicated :rolleyes:

wherever did x come from? :confused:

(and you never answered:
what is the number of ways of choosing 1 thing or 2 things or … n-1 things from n-1?
)
 

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