Let f : N --> R a function such that:

f(1) = 999 and
f(1) + f(2) + ... + f(n) = n²*f(n)

for all positive integer n.

Evaluate f(1998).

We know f(1)+...f(n) = n^2*f(n) and f(1)+...f(n+1)=(n+1)^2*f(n+1)

Thus, n^2*f(n)+f(n+1)=(n+1)^2*f(n+1). Subtracting f(n+1) from both sides gives n^2*f(n)=n*(n+2)*f(n+1). Dividing by n*(n+2) gives f(n+1)=f(n)*(n/(n+2))

Thus, f(n+2)=f(n)*(n*(n+1))/((n+2)*(n+3))... and by induction we can see f(n+k)=f(n) * (n*...*(n+k-1))/((n+2)*...(n+k+1))

Since we see each term from n+2 to n+k-1 appears on the top and bottom, cancelling those terms gives (n(n+1))/((n+k)(n+k+1))

Plugging in n=1, f(n)=999, and n+k=1998 gives 999*2/((1998)(1999))= 1999 as f(1998)

Posted by Gamer on 2008-10-31 15:19:08