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If $\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{n}{n+1} = \frac{1}{50}$, what is the sum of the numerator and denominator of the largest fraction on the left side of the equation?

 Nov 18, 2020
 #1
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productfor(n, 2, 99, n/(n+1)) = 1 / 50

 

The largest fraction =98 / 99

 

98 + 99 =197 sum of the numerator and denominator.

 Nov 18, 2020
 #2
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As follows:

\(\frac{2}{3}\frac{3}{4}...\frac{n}{n+1}=\frac{1}{50}\)  Succesive numerators and denominators cancel each other except for the very first numerator and the very last denominator, leaving \(\frac{2}{n+1}=\frac{1}{50}\)

 

Hence n+1 = 2*50   or n+1 = 100 and n = 99.  Sum = 199

 Nov 18, 2020

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