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?
 A: 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 $-\frac{1}{3}$.
A: This is how I would approach the problem. I don't know whether it's the most elegant solution, but to me it's the approach that stood out as the most immediate one to try. 
Note that $f(m, n)$ only depends on $n-m$. For instance, if $n-m = 1$, then $f(m,n) = \frac13$, and if $n-m = 2$, then there is only one $k$ with $m<k<n$, so $$f(m, n) = f(m, k) + f(k, n) - 2f(m, k)f(k, n) = \frac13 + \frac13 + 2\cdot \frac13\cdot\frac13 = \frac89$$
Now try with $n - m = 3$ and see what you get. See if you can spot a pattern, either at that point, or after $n-m = 4$ or $5$. With any luck, it'll be apparent by then.
A: 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
$$-\frac{1}{2}S = \frac{1}{6}$$
$$S = -\frac{1}{3}$$
A: 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.
