# fraction problems

• Mar 20th 2008, 11:02 PM
cdesiani
fraction problems

1.find the value of 1/(4x9) + 1/(9x14) + 1/(14x19) + ........ + 1/(1999x2004)

2.find the value of 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + ....... + 1/300

• Mar 21st 2008, 05:57 AM
teuthid
what level of math did this problem come from? Would it make sense to you if I told you that the first sum could be expressed as $\displaystyle\sum_{n=1}^{400}\frac{1}{(5n-1)(5n+4)}$?
• Mar 21st 2008, 07:50 AM
cdesiani
This problem was one of the problems in Math Olympiad for the elementary school. The time allocated for the problems is about 5 minutes. So there must be simple but logical answer. Please help...
(Crying)
• Mar 21st 2008, 08:40 AM
teuthid
Sorry, I can't think of anything that could be done by an elementary student in 5 minutes.
• Mar 21st 2008, 04:29 PM
Soroban
Hello, cdesiani!

Elementary school?
Perhaps very bright elementary students who have mastered Algebra
. . and know some clever (sneaky) tricks . . .

Quote:

1) Find the value of: . $\frac{1}{4\!\cdot\!9} + \frac{1}{9\!\cdot\!14} + \frac{1}{14\!\cdot\!19} + \cdots + \frac{1}{1999\!\cdot\!2004}$

We note that: . $\begin{Bmatrix}\dfrac{1}{4 \cdot 9} &=&\dfrac{1}{5}\left(\dfrac{1}{4} - \dfrac{1}{9}\right) \\ \\[-2mm]
\dfrac{1}{9\cdot14} &=&\dfrac{1}{5}\left(\dfrac{1}{9} - \dfrac{1}{14}\right) \\ \\[-2mm]
\dfrac{1}{14\cdot9} &=&\dfrac{1}{5}\left(\dfrac{1}{14} - \dfrac{1}{19}\right) \\ \vdots & & \vdots \end{Bmatrix}$

We have: . $\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}\right) + \frac{1}{5}\left(\frac{1}{9}-\frac{1}{14}\right) + \frac{1}{5}\left(\frac{1}{14} - \frac{1}{19}\right) + \hdots + \frac{1}{5}\left(\frac{1}{1999} - \frac{1}{2004}\right)$

. . . . . . $= \;\frac{1}{5}\left(\frac{1}{4} - \frac{1}{9} + \frac{1}{9} - \frac{1}{14} + \frac{1}{14} - \frac{1}{19} + \hdots + \frac{1}{1999} - \frac{1}{2004}\right)$

. . . . . . $= \;\frac{1}{5}\left(\frac{1}{4} - \frac{1}{2004}\right) \;=\;\frac{1}{5}\left(\frac{500}{2004}\right) \;=\;\boxed{\frac{25}{501}}$

Quote:

2) Find the value of: . $\frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + \cdots + \frac{1}{300}$

We note that: . $\begin{Bmatrix}3 &=&\dfrac{2\cdot3}{2} \\ \\[-2mm]
6 &=&\dfrac{3\cdot4}{2} \\ \\[-2mm]
10 &=&\dfrac{4\cdot5}{2} \\ \\[-2mm]
15 &=&\dfrac{5\cdot6}{2} \\ & \vdots \end{Bmatrix}$

We have: . $\frac{1}{\frac{2\cdot3}{2}} + \frac{1}{\frac{3\cdot4}{2}} + \frac{1}{\frac{4\cdot5}{2}} + \frac{1}{\frac{5\cdot6}{2}} + \hdots + \frac{1}{\frac{24\cdot25}{2}}$. . . . . $= \;\frac{2}{2\cdot3} + \frac{2}{3\cdot4} + \frac{2}{4\cdot5} + \frac{2}{5\cdot6} + \hdots + \frac{2}{24\cdot25}$

. . . . . $= \;2\left[\frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \frac{1}{4\cdot5} + \frac{1}{5\cdot6} + \hdots + \frac{1}{24\cdot25}\right]$

. . . . . $= \;2\left[\left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \hdots + \left(\frac{1}{24} - \frac{1}{25}\right)\right]$

. . . . . $= \;2\left(\frac{1}{2}-\frac{1}{25}\right) \;=\;2\left(\frac{23}{50}\right) \;=\;\boxed{\frac{23}{25}}$

By sixth grade I had mastered tying my shoelaces and finding my mittens.
.
• Mar 23rd 2008, 05:58 AM
cdesiani
Thank you, it's solved!!
Wow, what a beautiful way to solve such a difficult problem!

Thank you so much!
(Rofl)
• Apr 11th 2008, 01:08 PM
samlexii1403@yahoo.com
WELL........
I CAN NOT HELP U I DO NOT NO HOW 2 DO IT ETHER !!!!(Emo)

*****SORRY!!!!!!!!!!*****
• Apr 11th 2008, 01:09 PM
Moo
Hi,

Please be nice with my eyes (Crying)