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?
+23246 Notice that the 3 of the first fraction cancels the three of the second fraction;
that the 4 of the second fraction cancels the four of the third fraction;
that the 5 of the third fraction cancels the five of the fourth fraction;
etc.
This means that the only numbers remaining on the left side is the first numerator of 2 and the last denominator of n + 1.
The equation now becomes: 2 / (n + 1) = 1 / 50
Cross-multiplying: (2)(50) = (1)(n + 1)
100 = n + 1
99 = n
The largest fraction on the left-hand side is: 99/100
+23246 Notice that the 3 of the first fraction cancels the three of the second fraction;
that the 4 of the second fraction cancels the four of the third fraction;
that the 5 of the third fraction cancels the five of the fourth fraction;
etc.
This means that the only numbers remaining on the left side is the first numerator of 2 and the last denominator of n + 1.
The equation now becomes: 2 / (n + 1) = 1 / 50
Cross-multiplying: (2)(50) = (1)(n + 1)
100 = n + 1
99 = n
The largest fraction on the left-hand side is: 99/100
