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?
+23246 2/3 x 3/4 x 4/5 x 5/6 x ... x n/(n + 1) = 1 / 500
Notice that the denominator of the first fraction cancels the numerator of the second fraction and this pattern continues throughtout.
So, the only numbers left on the left-hand-side are 2 and n + 1.
This means that 2 / (n + 1) = 1 / 500 ---> 1000 = n + 1 ---> n = 999
So: n / (n + 1) = 999 / 1000
+23246 2/3 x 3/4 x 4/5 x 5/6 x ... x n/(n + 1) = 1 / 500
Notice that the denominator of the first fraction cancels the numerator of the second fraction and this pattern continues throughtout.
So, the only numbers left on the left-hand-side are 2 and n + 1.
This means that 2 / (n + 1) = 1 / 500 ---> 1000 = n + 1 ---> n = 999
So: n / (n + 1) = 999 / 1000
