Levans writes a positive fraction in which the numerator and denominator are integers, and the numerator is 1 greater than the denominator. He then writes several more fractions. To make each new fraction, he increases both the numerator and the denominator of the previous fraction by 1 . He then multiplies all his fractions together. He has 20 fractions, and their product equals 3 . What is the value of the first fraction he wrote?

I realize you need to do cancellation of some sort, but dont exactly know how to do it.

Thank you!

Guest May 23, 2022

#2**0 **

We have the equation: \({{x + 1} \over {x}} \times {x + 2 \over x+1} \times {x + 3 \over x + 2} \times \cdot \cdot \cdot \times {x+21 \over x+20} = 3\).

Note that the \(x+1\) all the way to the \(x+20\) terms cancel out.

This gives us: \({{x + 21} \over {x}} = 3\).

Can you take it from here?

BuilderBoi May 23, 2022

#4**0 **

Close...

We have the equation: \({{x + 21} \over x} = 3\).

Cross multiplying gives us: \(x + 21 = 3x \).

Subtracting x from both sides gives us: \(21 = 2x \)

This means \(x = {21 \over 2}\).

Now, we subsitute this in, and get: \(\large{{21 \over 2} + 1 \over {21 \over 2}}\).

Can you take it from here?

BuilderBoi
May 23, 2022

#6**0 **

Wait... my bad... I made an error. In my original answer, there are 21 terms, not 20.

It should be \({x+1 \over x} \times {x +2 \over x+1} \times {x +3 \over x +2 } \times \cdot \cdot \cdot \times {x + 20 \over x +19} = 3\).

This means we have the equation: \(x + 20 = 3x\), where x = 10, meaning the original fraction was \(\color{brown}\boxed{11 \over 10}\)

BuilderBoi
May 23, 2022