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Thread: combinatorics question

  1. Dec 6th, 2002, 11:47 PM #1
    bugzpodder
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    combinatorics question

    prove that (n_C_0)^2+(n_C_1)^2+...+(n_C_n)^2=2n_C_n

    (one more thing in case of Kalk, no induction )
    Massey RuleZ! ^-^__Cheers!__^-^ Massey RuleZ!


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  2. Dec 7th, 2002, 10:24 PM #2
    Kalkewl8ter
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    Wink

    Lol...why no induction, bugz?
    Merry Math Making!
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  3. Dec 8th, 2002, 09:40 AM #3
    bugzpodder
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    because it can be done without induction!

    (x+1)^(2n)=(x+1)^n * (x+1)*n

    the LHS has a term of 2n_C_n *x^n

    the RHS has x^n with coefficients (n_C_0)(n_C_n)+(n_C_1)(n_C_n-1)+...+(n_C_n)(n_C_1)

    so (n_C_0)=(n_C_n) and (n_C_1)=(n_C_n-1) and ...

    so therefore

    2n_C_n=Sum (r=0..n) n_C_r
    Massey RuleZ! ^-^__Cheers!__^-^ Massey RuleZ!


    Did you know that...
    The probability that a random rational number has an even denominator is 1/3 (Salamin and Gosper 1972)? This result is independently verified by me (2002)!
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  4. Dec 13th, 2002, 05:22 PM #4
    Kalkewl8ter
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    Ok bugz, I'm getting sick of your elegant solutions! You know way too much for the rest of us!
    Merry Math Making!
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  6. Dec 20th, 2002, 06:11 AM #5
    sql_lall
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    Talking Just a suggestion.

    Hi, just a trivial matter, but it may be easier to understand if you use HTML tags. i.e. instead of n_C_r, wriet nCr


    Oh, and I take it your last line was meant to say:
    (2n)_C_n=Sum (r=0..n) n_C_r ^2

    But anyway, its a nice proof.
    sql_lall
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