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Math Help - Binomial Theorem?

  1. #1
    Super Member fardeen_gen's Avatar
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    Binomial Theorem?

    Given that:
    S_{n} = 1 + q + q^2 + ... + q^n
    \sigma_{n} = 1 + \left(\frac{q + 1}{2}\right) + \left(\frac{q + 1}{2}\right)^2 + ... + \left(\frac{q + 1}{2}\right)^n
    q\neq1
    Prove that:
    {n + 1\choose 1} + {n + 1\choose 2}\cdot S_{1} + {n + 1\choose 3}\cdot S_{2} + ... + {n + 1\choose n + 1}\cdot S_{n} = 2^n\sigma_{n}
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    Given that:
    S_{n} = 1 + q + q^2 + ... + q^n
    \sigma_{n} = 1 + \left(\frac{q + 1}{2}\right) + \left(\frac{q + 1}{2}\right)^2 + ... + \left(\frac{q + 1}{2}\right)^n
    q\neq1
    Prove that:
    {n + 1\choose 1} + {n + 1\choose 2}\cdot S_{1} + {n + 1\choose 3}\cdot S_{2} + ... + {n + 1\choose n + 1}\cdot S_{n} = 2^n\sigma_{n}
    Hi fardeen_gen.

    Note that S_r=\frac{1-q^{r+1}}{1-q},\quad0\le r\le n.


    Hence \sum_{r\,=\,0}^n{n+1\choose r+1}S_r


    __ =\ \frac1{1-q}\sum_{r\,=\,0}^n{n+1\choose r+1}(1-q^{r+1})


    __ =\ \frac1{1-q}\sum_{r\,=\,0}^n{n+1\choose r+1}\,-\,\frac1{1-q}\sum_{r\,=\,0}^n{n+1\choose r+1}q^{r+1}


    __ =\ \frac1{1-q}\left[2^{n+1}-1\right]\,-\,\frac1{1-q}\left[(1+q)^{n+1}-1\right]


    __ =\ \frac{2^{n+1}-(1+q)^{n+1}}{1-q}


    __ =\ \frac{2^{n+1}\left[1-\left(\frac{1+q}2\right)^{n+1}\right]}{1-q}


    __ =\ \frac{2^n\left[1-\left(\frac{1+q}2\right)^{n+1}\right]}{\frac{1-q}2}


    __ =\ \frac{2^n\left[1-\left(\frac{1+q}2\right)^{n+1}\right]}{1-\frac{1+q}2}


    __ =\ 2^n\sigma_n
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