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Math Help - help me to simplify this

  1. #1
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    help me to simplify this

    \sum\limits_{n = 0}^{n - 1} \left({\frac {1}{2}}\right) ^ n
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  2. #2
    Super Member wingless's Avatar
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    I assume that n-1 on sigma is a constant and not related to the summation.

    \sum _{k=0}^a r^k = \frac{r^{1+a}-1}{r-1}

    So,
    \sum_{n = 0}^{n - 1} \left({\frac {1}{2}}\right)^n = \frac{\left(\frac{1}{2}\right)^{1+(n-1)}-1}{\frac{1}{2}-1} = 2 - 2^{1-n}
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  3. #3
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    wel wel wel

    well i think the answer to that question is 1 though im not sure
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  4. #4
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    Quote Originally Posted by gen2dan View Post
    well i think the answer to that question is 1 though im not sure
    If it was the infinite series \sum_{n=1}^{\infty} \left( \frac{1}{2} \right)^n then the answer would indeed be 1.

    But it's not so it's not.

    (Note: The very first term of the given series is \left( \frac{1}{2} \right)^0 = 1 so clearly it can't sum to 1 .......)
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