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 $m&lt;k&lt;n$, 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?
Note that:

$$ f(1,99)=f(1,k)+f(k,99)-2f(1,k) \cdot f(k,99) $$
and we may pick $k=98$:
$$ f(1,99)=f(1,98)+f(98,99)-2f(1,98) \cdot f(98,99) $$
As $f(98,99)=\frac{1}{3}$:
$$ f(1,99)=f(1,98)+\frac{1}{3}-2f(1,98) \cdot \frac{1}{3} $$
Let's simplify this a bit:
$$ f(1,99)=\frac{1}{3}f(1,98)+\frac{1}{3} $$
and rearrange:
$$ -\frac{1}{3}=\frac{1}{3}f(1,98)-f(1,99) $$
Thus the answer is $-\frac{1}{3}$.
Hint Try to prove that $\frac{1+f(m,m+n-1)}{f(m,m+n)}=3$ Then put m=1,n=99 to yield the expression you want.