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Math Help - Arithemetic sums

  1. #1
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    Arithmetic sums

    Dear forum members,


    please can someone help me to understand how to do this problem


    Determine the sum of all quotients \frac{m}{n} where m and n are whole numbers and 0<m<n<100.


    I don't understand how to do this.


    Any help is appreciated.





    Thank you in advance!
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  2. #2
    Flow Master
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    Quote Originally Posted by Coach View Post
    Dear forum members,


    please can someone help me to understand how to do this problem


    Determine the sum of all quotients \frac{m}{n} where m and n are whole numbers and 0<m<n<100.


    I don't understand how to do this.


    Any help is appreciated.





    Thank you in advance!
    \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ..... + \frac{1}{99}\right)

    + \left( \frac{2}{3} + \frac{2}{4} + \frac{2}{5} + ..... + \frac{2}{99}\right)

    + \left( \frac{3}{4} + \frac{3}{5} + \frac{3}{6} + ..... + \frac{3}{99}\right)

    + .....

    + \left( \frac{97}{98} + \frac{97}{99}\right)

    + \left( \frac{98}{99}\right).


    There is now a clever grouping you can do:

    \frac{1}{2} + \left( \frac{1}{3} + \frac{2}{3}\right) + \left( \frac{1}{4} + \frac{2}{4} + \frac{3}{4}\right) + ....... + \left( \frac{1}{99} + \frac{2}{99} + \frac{3}{99} + ..... + \frac{98}{99} \right )



    = \frac{1}{2} + \frac{1}{3} (1 + 2) + \frac{1}{4} (1 + 2 + 3) + ....... + \frac{1}{99} (1 + 2 + 3 + .... + 97 + 98)

    and each bracket contains a simple arithmetic series ...... Note: The sum of the first n natural numbers is given by \frac{n(n+1)}{2}.
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  3. #3
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    Thank you so much!


    Now I get the method. But won't it be quite a long calculation, even if using the arithmetic formula. My teacher wrote a "short cut" on a sample answer sheet as follows



    \sum^{99}_{i=2}=\frac{\sum^{i-1}_{k=1}=k}{i}=\sum^{99}_{i=2}=\frac{(i-1)*\frac{1+i-1}{2}}{i}


    the i-1 should be above the sigma sign, and the k=1 below, and there should be no equal sign before the k, but I just can't get it to work like that. Hope it is still clear enough.

    Can you please help me figure out what he is trying to tell?



    I noticed that the amount numbers in brackets after the fraction multiplier, are always one less in amount than the numerator of the fraction, but I don't know how to make use of that when creating a shortcut to calculate the sum.
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  4. #4
    Moo
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    Hello,

    Did you understood why he gave this formula ? I'll try to explain a part of it...

    This is because in a first time, you sum the fractions while variating the numerator, then when variating the denominator.

    The numerator is, as Mr F mentioned the sum of the terms of an arithmetic sequence.

    The thing is, normally, it's n(n+1)/2, with n the number above the sign sum (can't find a noun about it)

    So here, it's (i-1)(i-1+1)/2 according to the formula.
    But i-1+1=i
    So you can simplify the general term of the sequence.

    And you now have to calculate \sum_{i=2}^{99} \frac{i-1}{2} = \frac{1}{2} (\sum_{i=2}^{99} i) - \frac{98}2
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  5. #5
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    We all have different ways. This is close to your instructor's.

    \sum\limits_{n = 2}^{99} {\frac{{\sum\nolimits_{m = 1}^{n - 1} m }}{n}}  = \sum\limits_{n = 2}^{99} {\frac{{\frac{{\left( {n - 1} \right)n}}{2}}}{n}}  = \frac{1}{2}\sum\limits_{n = 2}^{99} {(n - 1)}  = \frac{1}{2}\sum\limits_{n = 1}^{98} n  = \frac{{\left( {98} \right)\left( {99} \right)}}{4}<br />
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