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?
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
+2666 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 \(999 + 1000 =\color{brown}\boxed{1,999}\)
