A hard fraction problem

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August 19th 2012, 03:53 AM
Notatroll

A hard fraction problem

I have a problem with a fraction.

(1+1/1)*(1+1/2)*(1+1/3)…(1+1/99)

How should i solve this?

Thx in advance
August 19th 2012, 04:29 AM
emakarov

Re: A hard fraction problem

Hint: express each factor as an improper fraction.
August 19th 2012, 05:19 AM
Notatroll

Re: A hard fraction problem

Can you give an example cuz I'm not good at these terms in english.
August 19th 2012, 05:33 AM
Plato

Re: A hard fraction problem

Quote:

Originally Posted by Notatroll

Can you give an example cuz I'm not good at these terms in english.

$\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2} \right)\left(1+\frac{1}{3}\right)\cdots\left(1+ \frac{1}{99}\right)=\left(\frac{2}{1}\right)\left( \frac{3}{2}\right)\left(\frac{4}{3}\right)\cdots \left(\frac{100}{99}\right)$
August 19th 2012, 05:41 AM
Notatroll

Re: A hard fraction problem

Yeah I hae done that but how am I supposed to solve it? It's like 100000000 digits long :S
August 19th 2012, 06:24 AM
Notatroll

Re: A hard fraction problem

Oh thx. I just got it, isn't it f(x)=x+2? Every time you multiplice it it grows with 1, 98 times. Thx again to you Plato and emakarov!
August 19th 2012, 08:12 AM
emakarov

Re: A hard fraction problem

Quote:

Originally Posted by Notatroll

Can you give an example cuz I'm not good at these terms in english.

I also had to look it up in Wikipedia before I replied. Is it too difficult to Google it?

Quote:

Originally Posted by Notatroll

Oh thx. I just got it, isn't it f(x)=x+2?

I don't know what you mean by f(x).

Quote:

Originally Posted by Notatroll

Every time you multiplice it it grows with 1, 98 times.

This is also not clear. When you multiply, say (1 + 1/1)(1 + 1/2) by (1 + 1/3), it does not grow by 1.
August 19th 2012, 08:51 AM
Plato

Re: A hard fraction problem

Quote:

Originally Posted by Notatroll

Yeah I hae done that but how am I supposed to solve it? It's like 100000000 digits long :S

$\left( {\frac{{\rlap{/} 2}}{1}} \right)\left( {\frac{{\rlap{-/} 3}}{{\rlap{/} 2}}} \right)\left( {\frac{{\rlap{/} 4}}{{\rlap{/} 3}}} \right) \cdots \left( {\frac{{\rlap{/} 9\rlap{/} 9}}{{\rlap{/} 9\rlap{/} 8}}} \right)\left( {\frac{{100}}{{\rlap{/} 9\rlap{/} 9}}} \right) = 100$
August 19th 2012, 08:21 PM
richard1234

Re: A hard fraction problem

Rewrite $1 + \frac{1}{1}$ as $\frac{2}{1}$ , $1 + \frac{1}{2}$ as $\frac{3}{2}$ and so on. Pretty much everything cancels, as Plato showed. It essentially "telescopes."
August 26th 2012, 10:33 AM
Notatroll

Re: A hard fraction problem

Ehh... (2/1)*(3/2)*(4/3) is actually 4, so it does grow with 1 all the time.. So i suppose i was correct.