The number of functions

$f:\{ 1,2,3,4\} \to \{ a \in Z|a| \le 8\}$

satisfying $f(n) + {1 \over n}f(n + 1) = 1,\forall n \in \{ 1,2,3\}$ is

$\because f:\{1,2,3,4\} \rightarrow\{a \in \mathbb{Z}:|9| \leq 8\}$ <br/><br/> and $f(n)+\frac{1}{n} f(n+1)=1$ <br/><br/> $\Rightarrow n f(n)+f(n+1)=n \quad \ldots$ (i) <br/><br/> $\therefore f(1)+f(2)=1 \Rightarrow f(2)=1-f(1)$ <br/><br/> But $f(1) \in[-8,8]$ <br/><br/> Hence, $f(2) \in[-8,8] \Rightarrow f(1) \in[-7,8] \quad\ldots(\mathrm{A})$ <br/><br/> and $2 f(2)+f(3)=2 \Rightarrow f(3)=2 f(1)$ <br/><br/> $\therefore 2 f(1) \in[-8,8] \Rightarrow f(1) \in[-4,4] \quad\ldots(\mathrm{B})$ <br/><br/> and $3 f(3)+f(4)=3 \Rightarrow f(4)=3-6 f(1)$ <br/><br/> $\therefore f(1) \in\left[-\frac{5}{6}, \frac{11}{6}\right]\quad...(C)$ <br/><br/> From (A), (B) and (C) : $f(1)=0$ or 1 <br/><br/> $\therefore$ Only two functions are possible.