# Math Help - A hard fraction problem

1. ## 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?

2. ## Re: A hard fraction problem

Hint: express each factor as an improper fraction.

3. ## Re: A hard fraction problem

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

4. ## Re: A hard fraction problem

Originally Posted by Notatroll
Can you give an example cuz I'm not good at these terms in english.
$\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2} \right)\left(1+\frac{1}{3}\right)\cdots\left(1+ \frac{1}{99}\right)=\left(\frac{2}{1}\right)\left( \frac{3}{2}\right)\left(\frac{4}{3}\right)\cdots \left(\frac{100}{99}\right)$

5. ## 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

6. ## 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!

7. ## Re: A hard fraction problem

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?

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).

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.

8. ## Re: A hard fraction problem

Originally Posted by Notatroll
Yeah I hae done that but how am I supposed to solve it? It's like 100000000 digits long :S
$\left( {\frac{{\rlap{/} 2}}{1}} \right)\left( {\frac{{\rlap{-/} 3}}{{\rlap{/} 2}}} \right)\left( {\frac{{\rlap{/} 4}}{{\rlap{/} 3}}} \right) \cdots \left( {\frac{{\rlap{/} 9\rlap{/} 9}}{{\rlap{/} 9\rlap{/} 8}}} \right)\left( {\frac{{100}}{{\rlap{/} 9\rlap{/} 9}}} \right) = 100$

9. ## 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."

10. ## 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.