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

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- Aug 19th 2012, 03:53 AMNotatrollA 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 - Aug 19th 2012, 04:29 AMemakarovRe: A hard fraction problem
Hint: express each factor as an improper fraction.

- Aug 19th 2012, 05:19 AMNotatrollRe: A hard fraction problem
Can you give an example cuz I'm not good at these terms in english.

- Aug 19th 2012, 05:33 AMPlatoRe: A hard fraction problem
- Aug 19th 2012, 05:41 AMNotatrollRe: A hard fraction problem
Yeah I hae done that but how am I supposed to solve it? It's like 100000000 digits long :S

- Aug 19th 2012, 06:24 AMNotatrollRe: 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!

- Aug 19th 2012, 08:12 AMemakarovRe: A hard fraction problem
I also had to look it up in Wikipedia before I replied. Is it too difficult to Google it?

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

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. - Aug 19th 2012, 08:51 AMPlatoRe: A hard fraction problem
$\displaystyle \left( {\frac{{\rlap{/} 2}}{1}} \right)\left( {\frac{{\rlap{-/} 3}}{{\rlap{/} 2}}} \right)\left( {\frac{{\rlap{/} 4}}{{\rlap{/} 3}}} \right) \cdots \left( {\frac{{\rlap{/} 9\rlap{/} 9}}{{\rlap{/} 9\rlap{/} 8}}} \right)\left( {\frac{{100}}{{\rlap{/} 9\rlap{/} 9}}} \right) = 100$

- Aug 19th 2012, 08:21 PMrichard1234Re: A hard fraction problem
Rewrite $\displaystyle 1 + \frac{1}{1}$ as $\displaystyle \frac{2}{1}$, $\displaystyle 1 + \frac{1}{2}$ as $\displaystyle \frac{3}{2}$ and so on. Pretty much everything cancels, as Plato showed. It essentially "telescopes."

- Aug 26th 2012, 10:33 AMNotatrollRe: 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.