It's Mathematics Olympiad! - 1

$\begin{aligned} f(1) &= 1^2 f(1) \\ f(1)+f(2) &= 2^2 f(2) \implies f(2) = \cfrac{1}{3}f(1)\\ f(1)+f(2)+f(3) &= 3^2 f(3) \implies f(3) = \cfrac{4}{8}f(2) = \cfrac{4}{8}\cfrac{1}{3} f(1) \\ f(1)+f(2)+f(3)+f(4) &= 4^2 f(4) \implies f(4) = \cfrac{9}{15}f(3) = \cfrac{9}{15}\cfrac{4}{8}\cfrac{1}{3} f(1)\\ \sum\limits_{k=1}^{n} f(k) - \sum\limits_{k=1}^{n-1} f(k) &= n^2 f(n) - (n-1)^2 f(n-1) \\ \implies f(n) &= n^2 f(n) - (n-1)^2 f(n-1) \\ \implies f(n) &= \cfrac{(n-1)^2}{n^2-1} f(n-1) = \cfrac{(n-1)}{n+1} f(n-1) \\ \end{aligned}$

So, we have, for $n\geq 2$

$f(n) = \prod\limits_{k=2}^{n} \cfrac{n-1}{n+1} f(1) = \cfrac{2 f(1)}{(n+1) n}$

$\therefore f(1996) = \cfrac{\cancel{1995}}{1997} \cfrac{\cancel{1994}}{1996} \cfrac{\cancel{1993}}{\cancel{1995}} ...\cfrac{2}{\cancel{4}}\cfrac{1}{\cancel{3}} f(1) = \cfrac{2 \times 1996}{1997\times1996} = \cfrac{2}{1997}$

Hence, $a+b = 2+1997 = \boxed{1999}$

Observation :

$\boxed{Conjecture : f(n)=\dfrac{1996}{\frac{n}{2}(n+1)}=\dfrac{1996\times 2}{n(n+1)}}$ $Proof :$ $f(1)=\dfrac{1996\times \cancel{2}}{1\times \cancel{(1+1)}}=1996..........[1]$ $\sum_{k=1}^nf(k)=\sum_{k=1}^n\dfrac{1996\times 2}{k(k+1)}$ $1996\times 2\sum_{k=1}^n\dfrac{(k+1)-k}{k(k+1)}$ $1996\times 2\sum_{k=1}^n\dfrac{\cancel{k+1}}{k\cancel{(k+1)}}-1996\times 2\sum_{k=1}^n\dfrac{\cancel{k}}{\cancel{k}(k+1)}$ $=1996\times 2\sum_{k=1}^n\dfrac{1}{k}-1996\times 2\sum_{k=1}^n\dfrac{1}{k+1}$ $=1996\times 2\sum_{k=1}^n\dfrac{1}{k}-1996\times 2\sum_{k=2}^{n+1}\dfrac{1}{k}$ $=1996\times 2(1-\dfrac{1}{n+1}\cancel{\sum_{k=2}^n\dfrac{1}{k}-\sum_{k=2}^{n}\dfrac{1}{k}})$ $=1996\times \dfrac{2n}{n+1}$ $=1996\times \dfrac{2n\red{\times n}}{(n+1)\red{\times n}}$ $=1996\times \dfrac{2n^2}{(n+1)n}$ $=n^2\times\red{\dfrac{1996\times 2}{n(n+1)}}$ $=n^2f(n)$ $Q.E.D.$ $f(1996)=\dfrac{\cancel{1996}\times 2}{\cancel{1996}\times(1996+1)}=\dfrac{2}{1997}$ $\because \gcd(2,1997)=1$ $\therefore \boxed{Answer=1997+2=1999}$