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Math Help - Proof Help!!

  1. #1
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    Proof Help!!

    Prove that:

    "The summation" from k=2 to n of (k choose 2) = (n+1 choose 3). Let n be a natural number.

    Please Help. thanks!
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  2. #2
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    Hello, jzellt!

    It helps if we know a few summation formulas:

    . . \sum^n_{k=1} k \;=\;\frac{k(k+1)}{2}\qquad\quad \sum^n_{k=1}k^2 \;=\;\frac{n(n+1)(2n+1)}{6}


    Prove that: . \sum^n_{k=2}{k\choose2} \;=\;{n+1\choose3}

    \sum^n_{k=2}{k\choose2} \;=\;{2\choose2} + {3\choose2} + {4\choose2} + {5\choose2} + \hdots + {n\choose2}

    . . . . . . = \;1 + 3 + 6 + 10 + \hdots + \frac{n(n-1)}{2}

    This is the sum of the first n-1 triangular numbers.


    We have: . \sum^{n-1}_{k=1}\frac{k(k+1)}{2} \;\;=\;\;\frac{1}{2}\bigg[\sum^{n-1}_{k=1}k^2 + \sum^{n-1}_{k=1} k\bigg] \;\;=\;\;\frac{1}{2}\bigg[\frac{n(n-1)n(2n-1)}{6} + \frac{n(n-1)}{2}\bigg]

    Factor: . \frac{n(n-1)}{12}(2n-1 + 3) \;\;=\;\;\frac{n(n-1)}{12}(2n+2) \;\;=\;\;\frac{n(n-1)}{12}\cdot2(n+1)


    And we have: . \frac{(n+1)n(n-1)}{3!}\quad\hdots\quad\text{which equals: }\:{n+1\choose3}

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