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TrojanPoem

  • one year ago

Prove that: [(2n+1)C(0)]^2 - [(2n+1)C(1)]^2 + [(2n+1)C(2)]^2 -..... - ([(2n+1)C(2n+1)]^2 = 0

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  1. TrojanPoem
    • one year ago
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    Someone rewrite it with EQN I can't.

  2. dan815
    • one year ago
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    like thsi right?

  3. dan815
    • one year ago
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    |dw:1435164304284:dw|

  4. TrojanPoem
    • one year ago
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    Use another method this E is out of my curriculum.

  5. dan815
    • one year ago
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    it just means add

  6. dan815
    • one year ago
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    but i think this ons is pretty straight forawrd because

  7. dan815
    • one year ago
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    we have symmetry along the pascal triangle

  8. TrojanPoem
    • one year ago
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    You know , it's meant to be proved with binomial

  9. dan815
    • one year ago
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    that is binomial

  10. dan815
    • one year ago
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    [(2n+1)C(0)]^2 = [(2n+1)C(2n+1)]^2 [(2n+1)C(1)]^2=[(2n+1)C(2n)]^2 [(2n+1)C(2)]^2=[(2n+1)C(2n-1)]^2 and so on right

  11. dan815
    • one year ago
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    im not satisfied with this though something is not making sense let me check something

  12. dan815
    • one year ago
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    |dw:1435164616647:dw|

  13. dan815
    • one year ago
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    ok silly yes it works xD

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