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
$$\frac{1}{3} f(1,98)-f(1,99)$$
Could someone give me hint as how to initiate this problem?
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.
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 $-\frac{1}{3}$.
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
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
$$-\frac{1}{2}S = \frac{1}{6}$$ $$S = -\frac{1}{3}$$