let $f(1)=c=\frac{2c}2; \implies f(2)+f(1)=4f(2)\implies f(2)=\frac{c}3=\frac{2c}6$ Let prove then, $f(n)=\frac{2f(1)}{n(n+1)}$ by induction on n; We have the base of the induction. for $f(1), f(2)$; let it is true for $n: f(n)=\frac{2c}{n(n+1)} \implies f(n+1)+f(n)+\dots +f(1)=(n+1)^2f(n+1)$ $$f(n+1)+\frac{2c}{n(n+1)}+\frac{2c}{n(n-1)}+\dots + \frac{2c}{1\cdot 2}=(n+1)^2f(n+1)$$$$\sum_{i=1}^{n}\frac{2c}{i(i+1)}=2c\sum_{i=1}^{n}\frac 1{i(i+1)}=2c\left( \frac 1{n(n+1)} + \dots \frac 12 \right)=2c\left( \frac 1{n}-\frac 1{n+1}+\frac 1{n-1} -\frac 1{n} +\dots \frac 12 -\frac 13 +\frac 11 -\frac 12 \right)= 2c \left(\frac 11 -\frac 1{n+1} \right )=2c\left(\frac{n}{n+1}\right) \implies$$$f(n+1)\left(n^2+2n \right)=\frac {2cn}{n+1} \implies f(n+1)=\frac {2c}{(n+1)(n+2)}$ hence, $\boxed{f(1995)=\frac{2c}{1996*1995}=\frac 2{3 982 020}}$