If function $f$ satisfies given conditions, then find $\frac{1}{3} f(1,98)-f(1,99)$ Let $f$ be a function defined on $\{(m,n):$ $m$ and $n$ are positive integers $\}$ satisfying:
$$1. f(m,m+1)=\frac{1}{3}$$, for all m


*$$f(m,n)=f(m,k)+f(k,n)-2f(m,k) \cdot f(k,n)$$ for all $k$ such that $m<k<n$, then find the value of 


$$\frac{1}{3} f(1,98)-f(1,99)$$
Could someone give me hint as how to initiate this problem?

First note the factor of $\frac{1}{3}$ and try to exploit it with the first defining equation:
$$\begin{align}S &= \frac{1}{3}f(1, 98) - f(1, 99)\\
&= f(98, 99)f(1, 98) - f(1, 99)\\
&= f(1, 98)f(98, 99) - f(1, 99)\end{align}$$
Let $m = 1, n = 99, k = 98$. Then, 
$$\begin{align}S &=\frac{f(1,98) + f(98, 99) - f(1, 99)}{2} - f(1, 99)\\
&= \frac{1}{2}f(1, 98) - \frac{3}{2}f(1, 99) + \frac{1}{6}\\
&= \frac{3}{2}\left(\frac{1}{3}f(1, 98) - f(1, 99)\right) + \frac{1}{6}\\
&= \frac{3}{2}S + \frac{1}{6}\end{align}$$
so that
$$-\frac{1}{2}S = \frac{1}{6}$$
$$S = -\frac{1}{3}$$
