Sum Square Function. Topic: Algebra/Sequences & Series. Level: Olympiad.

Problem: A function $f$ is defined for all positive integers and satisfies $f(1)=1996$, and $f(1)+f(2)+\cdots+f(n)=n^2f(n)$ for all $n>1$. Calculate the exact value of $f(1996)$.

Solution: Checking small values, we find $ f(2) = \frac{1996}{3} $, $ f(3) = \frac{1996}{6} $, and $ f(4) = \frac{1996}{10} $. The denominators look awfully like the triangular numbers, so we claim that

$ f(n) = \frac{2 \cdot 1996}{n(n+1)} $.

We will prove the result by strong induction.

Base Case - $ f(1) = 1996 $.

Induction Step - Assume that $ f(k) = \frac{2 \cdot 1996}{k(k+1)} $ for $ k = 1, \ldots, m $. We want to show that $ f(m+1) = \frac{2 \cdot 1996}{(m+1)(m+2)} $. From the given condition we have

$ f(1)+f(2)+\cdots+f(m)+f(m+1) = (m+1)^2f(m+1) $

$ \frac{2 \cdot 1996}{1 \cdot 2}+\frac{2 \cdot 1996}{2 \cdot 3}+\frac{2 \cdot 1996}{m(m+1)} = m(m+2)f(m+1) $.

Note that $ \frac{1}{k(k+1)} = \frac{1}{k}-\frac{1}{k+1} $, so the equation becomes

$ 2 \cdot 1996 \left( \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{m}-\frac{1}{m+1} \right) = m(m+2)f(m+1) $

$ 2 \cdot 1996 \left(1-\frac{1}{m+1}\right) = m(m+2)f(m+1) $

$ 2 \cdot 1996 \left(\frac{m}{m+1}\right) = m(m+2)f(m+1) $,

which rearranges to

$ f(m+1) = \frac{2 \cdot 1996}{(m+1)(m+2)} $,

completing the induction. Hence $ f(1996) = \frac{2 \cdot 1996}{1996 \cdot 1997} = \frac{2}{1997} $. QED.

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Comment: Usually problems like these require you to play around with the problem for a while and once you think you see a pattern, conjecture it and hopefully prove it. Induction works well especially when they ask for a large value such as $ 1996 $.

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Practice Problem: If $ f(1) = k $, evaluate $ f(k) $ with the same condition in the above problem.