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).

f(1)=999

now

f(1)+f(2)+...+f(n)=n^2*f(n)

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

so we have

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

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

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

(n-1)*f(n-1)=(n+1)*f(n)

f(n)=(n-1)*f(n-1)/(n+1)

so

f(1)=999

f(2)=f(1)/3

f(3)=2*f(2)/4=2*f(1)/12=f(1)/6

f(4)=3*f(3)/5=(3*2)*f(1)/(3*4*5)=f(1)/10

so in seems that it follows the patern

f(n)=2*f(1)/(n^2+n)

now to prove this using induction

f(1)=2*f(1)/(1+1)=2*f(1)/2=f(1)

thus it holds for initial value

assume it holds for f(n-1)

from definition derived above

f(n)=(n-1)*f(n-1)/(n+1)

substitute f(n-1)=2*f(1)/[n*(n-1)]

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

(n-1)'s drop out and we are left with

f(n)=2*f(1)/(n*(n+1))

thus if it holds for n-1 it holds for n

thus it holds for all n>=1

Posted by Daniel on 2008-10-31 14:36:05