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Math Help - Difficulty to understand the value of a sum

  1. #1
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    Difficulty to understand the value of a sum

    I am trying to understant the formula below, it is part of bigger exercise but I am stack on this partiular one:

    Sum from k=v+1 to n of (1/ (1/v+1)^(k-v-1)) = (1-(1/(v+1))^(n-v))/1-1/(v+1)
    where v<n

    This is in latex:
    \sum_{k=v-1}^{n}\frac{1}{(v+1)^{k-v-1}}=\frac{1-(\frac{1}{v+1})^{n-v}}{1-\frac{1}{v+1}}

    Thank you in advance!


    Sorry for latex, I am trying about 2 hours now.. The code produced from the buttons below doesn't work, the code in the latex forum doesn't work either, so I removed the math tags. Image upload does not work too, sorry!
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  2. #2
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    Quote Originally Posted by Melsi View Post
    I am trying to understant the formula below, it is part of bigger exercise but I am stack on this partiular one:

    Sum from k=v+1 to n of (1/ (1/v+1)^(k-v-1)) = (1-(1/(v+1))^(n-v))/1-1/(v+1)
    where v<n

    This is in latex:
     \sum_{k=v-1}^{n}\frac{1}{(v+1)^{k-v-1}}=\frac{1-(\frac{1}{v+1})^{n-v}}{1-\frac{1}{v+1}}
    LaTeX fix
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  3. #3
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    There is a sign error in the LaTeX above.
     \sum_{k=v+1}^{n}\frac{1}{(v+1)^{k-v-1}}=\frac{1-(\frac{1}{v+1})^{n-v}}{1-\frac{1}{v+1}} NOTE the k=v+1.

    Here is what you must know: \sum\limits_{k = A}^B {r^k }  = \frac{{r^A  - r^{B + 1} }}{{1 - r}},\,A < B

    Notice that: \sum\limits_{k = v + 1}^n {\left[ {\frac{1}{{v + 1}}} \right]^{k - v - 1} }  = \left[ {\frac{1}{{v + 1}}} \right]^{ - v - 1} \sum\limits_{k = v +1}^n{\left[{\frac{1}{{v + 1}}} \right]^k }

    So r=\left[ {\frac{1}{{v + 1}}} \right],~A=v+1,~\&~B=n

    Now you carry out the substitutions.
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  4. #4
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    Solved

    Thank you very much, it was the piece of information that I was missing! Glad to have it solved, I am applying the solution right away. Thank you for the latex help too!

    Bye!
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