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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... 23×34×45× ...=1500

 

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: 2x=1500 and x=1000

 

Using the infor given, the final fraction must be 9991,000, meaning the sum is 999+1000=1,999

 Mar 8, 2022

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