Manhattan challenge problem of the week! - June 12th

If the reciprocals of two consecutive integers are added to one another, what is the sum in terms of the greater integer x?

(A) 3/x
(B) x^2 – x
(C) 2x – 1
(D) 2x – 1 /x^2 + x
(E) 2x – 1 /x^2 – x

Solution:Official answer will be known next week - i.e June 19th

Answer must be E

Greater integer = x
Consecutive smaller integer number to x = x-1
Sum of reciprocals
=1/x +1/x-1 = x-1+x/x(x-1) = 2x - 1/(x^2 - x)

Thus E is the answer.