2012/034) Given f(1) + f(2) + ... + f(n) = (n^2)*[f(n)] f(1) = 999 Find f(1998)

From the given condition
f(1) + f(2) + ... + f(n) = (n^2)*[f(n)] ...1

f(1) + f(2) + ... + f(n-1) = (n-1)^2)*[f(n-1)] ..2

subtract (2) from (1)

f(n) = n^2f(n) - (n-1)^2 f(n-1)

or f(n) = (n-1)^2/(n^2-1) f(n-1) = (n-1)/(n+1) f(n-1)

f(2) = 1/3 f(1)
f(3) = 2/4 f(2) = 2f(1) / ( 3 * 4)
f(4) = 3/5 f(3)

from the above f(n) = 2f(1)/(n * (n+1))

so f(1998) = 2 * 999/( 1998 *1999) = 1/1999