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?
productfor(n, 2, 99, n/(n+1)) = 1 / 50
The largest fraction =98 / 99
98 + 99 =197 sum of the numerator and denominator.
+33616 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
