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y2o2

  • 2 years ago

prove that :

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  1. y2o2
    • 2 years ago
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    \[\large {^{2n} C _n} = {(^n C _0)}^2 +{(^n C _1)}^2 +{(^n C _2)}^2 +...+{(^n C _n)}^2 \]

  2. KingGeorge
    • 2 years ago
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    So in summation notation, show that\[\binom{2n}{n}=\sum_{i=0}^n \binom{n}{i}^2\]

  3. shubhamsrg
    • 2 years ago
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    you can use induction..

  4. Zarkon
    • 2 years ago
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    Use Vandermonde's identity

  5. Mimi_x3
    • 2 years ago
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    well have you tried it?

  6. Mimi_x3
    • 2 years ago
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    since it was from 2 months ago..

  7. y2o2
    • 2 years ago
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    Yes , and it came up with nothing

  8. vamgadu
    • 2 years ago
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    no it works.

  9. Mimi_x3
    • 2 years ago
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    \[LHS: \binom{2n}{n} x^{n} \] \[RHS\left[\binom{n}{0}\binom{n}{n}+\binom{n}{1}\binom{n}{n-1}+\binom{n}{2}\binom{n}{n-2}+....+\binom{n}{n-2}\binom{n}{2}+\binom{n}{n-1}\binom{n}{1}+\binom{n}{n}\binom{n}{0}\right] *x^n\] \[By ~Symmetry: \binom{n}{n} = \binom{n}{0} ; \binom{n}{n-1} = \binom{n}{1} \] \[hence~ RHS = \left[\binom{n}{0}^{2}+\binom{n}{1}^{2}+\binom{n}{2}^{2}+...+\binom{n}{n}^{2}\right] x^n\] \[hence, coeeficients~ of~ x^n~ are~ the ~same \] \[\therefore, \binom{n}{0}^{2}+\binom{n}{1}^{2}+\binom{n}{2}^{2}+...+\binom{n}{n}^{2} = \binom{2n}{n} \]

  10. vamgadu
    • 2 years ago
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    use this. its better. http://upload.wikimedia.org/wikipedia/en/math/4/0/6/406c9e30b2566d50df58a5dc3315d1d2.png

  11. Mimi_x3
    • 2 years ago
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    whats wrong with the method i used?

  12. vamgadu
    • 2 years ago
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    nothing. I said its easier to use vandermonde's identity.

  13. Mimi_x3
    • 2 years ago
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    okay; well i dont know how to use that identity lol

  14. vamgadu
    • 2 years ago
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    just plug in n every where except at k(its a summation variable).

  15. y2o2
    • 2 years ago
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    thank you, i got it now :)

  16. Mimi_x3
    • 2 years ago
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    You're welcome (:

  17. Mimi_x3
    • 2 years ago
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    Sorry I made a typo at the end: \[\therefore, \binom{n}{0}^{2}+\binom{n}{1}^{2}+\binom{n}{2}^{2}+...+\binom{n}{n}^{2} = \binom{2n}{n} ^2\]

  18. y2o2
    • 2 years ago
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    yeah , i realized that :)

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