Arithemetic sums

• Mar 29th 2008, 05:09 AM
Coach
Arithmetic sums
Dear forum members,

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

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

I don't understand how to do this.

Any help is appreciated.

Thank you in advance!
• Mar 29th 2008, 05:25 AM
mr fantastic
Quote:

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 $\displaystyle \frac{m}{n}$ where m and n are whole numbers and $\displaystyle 0<m<n<100.$

I don't understand how to do this.

Any help is appreciated.

Thank you in advance!

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

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

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

$\displaystyle + .....$

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

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

There is now a clever grouping you can do:

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

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

and each bracket contains a simple arithmetic series ...... Note: The sum of the first n natural numbers is given by $\displaystyle \frac{n(n+1)}{2}$.
• Mar 29th 2008, 06:24 AM
Coach
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

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

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.
• Mar 29th 2008, 07:20 AM
Moo
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 $\displaystyle \sum_{i=2}^{99} \frac{i-1}{2} = \frac{1}{2} (\sum_{i=2}^{99} i) - \frac{98}2$
• Mar 29th 2008, 09:00 AM
Plato
We all have different ways. This is close to your instructor's.

$\displaystyle \sum\limits_{n = 2}^{99} {\frac{{\sum\nolimits_{m = 1}^{n - 1} m }}{n}} = \sum\limits_{n = 2}^{99} {\frac{{\frac{{\left( {n - 1} \right)n}}{2}}}{n}} = \frac{1}{2}\sum\limits_{n = 2}^{99} {(n - 1)} = \frac{1}{2}\sum\limits_{n = 1}^{98} n = \frac{{\left( {98} \right)\left( {99} \right)}}{4}$