# Math Help - Arithemetic sums

1. ## 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

I don't understand how to do this.

Any help is appreciated.

2. Originally Posted by Coach
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

I don't understand how to do this.

Any help is appreciated.

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

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

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

$+ .....$

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

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

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 $\frac{n(n+1)}{2}$.

3. 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.

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.

4. 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 $\sum_{i=2}^{99} \frac{i-1}{2} = \frac{1}{2} (\sum_{i=2}^{99} i) - \frac{98}2$

5. 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}
$