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| f(1998) (Posted on 2008-10-31) |
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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).
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Submitted by pcbouhid
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Rating: 3.0000 (1 votes)
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Solution:
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Evaluating some values of f(n):
f(1) = 999
f(1) + f(2) = 22 * f(2) ---> 3*f(2) = 999 ---> f(2) = 333
f(1) + f(2) + f(3) = 32*f(3) ---> 8*f(3) = 999 + 333 ---> f(3) = 333/2
The same way, f(4) = 999/10
Thus, we have:
f(1) = 999/1, f(2) = 999/3, f(3) = 999/6, f(4) = 999/10.
Letīs prove by induction that f(n) = 999 / (1 + 2 + 3 + ... + n) = 999 / n(n+1)
For n >= 2, we have:
f(1) + f(2) + ... + f(n) = n2*f(n)
(n2 - 1)*f(n) = f(1) + f(2) + ... + f(n-1)
f(n) = [f(1) + f(2) + ... + f(n-1)] / (n2 - 1)
By the hypothesis of induction:
f(k) = 1998 / k(k + 1) = 1998/k - 1998/(k+1), for k = 1, 2, 3,..., (n-1), and so:
f(1) + f(2) + ... + f(n-1) = 1998/1 - 1998/2 + 1998/2 - 1998/3 + 1993/3 - 1998/4 ... + 1998/(n-1) - 1998/n =
= 1998/1 - 1998/n
f(n) = 1998(n-1)/n*(n2 - 1) = 1998 / n(n+1), as we wanted to prove.
Thus, making n = 1998 we find f(1998) = 1998 / (1998*1999)
f(1998) = 1/1999.
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