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If 2/3*3/4*4/5* ... * n/(n+1) = 1/500, what is the sum of the numerator and denominator of the largest fraction on the left side of the equation?

 Mar 7, 2022
 #1
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 2/3*3/4*4/5* ... * n/(n+1) = 1/500

Solve for n:
2/(n + 1) = 1/500

Take the reciprocal of both sides:
(n + 1)/2 = 500

Multiply both sides by 2:
n + 1 = 1000

Subtract 1 from both sides:

n = 999   and (n +1) ==1,000

 Mar 7, 2022
 #2
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The equation is... \({2 \over 3 } \times {3 \over 4 } \times { 4 \over 5} \times \space ... = {1 \over 500} \)

 

Notice that the 3s and 4s cancel out. The only number that will not get canceled is the denominator of the last fraction and the numerator of the first fraction (2).

 

This means: \({2 \over x} ={ 1 \over 500}\) and \(x = 1000\)

 

Using the infor given, the final fraction must be \(999 \over 1,000\), meaning the sum is \(999 + 1000 =\color{brown}\boxed{1,999}\)

 Mar 8, 2022

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