the range is given in this way:
y1=-1
y(n+1)=1/3y(n)+3
Find a f function y(n)=f(n)
thanks
For $\displaystyle y_{n+1} = \frac{1}{3} y_n + 3$ for let $\displaystyle y_n = z_n + k $ so the difference equation becomes
$\displaystyle z_{n+1} + k = z_n + \frac{1}{3}\left( z_n + k \right) + 3$
and choosing $\displaystyle k = \frac{9}{2}$ gives
$\displaystyle z_{n+1} = \frac{1}{3} z_n$
Seeking a solution of this in the form
$\displaystyle z_n = c \rho^n$ gives $\displaystyle \rho = \frac{1}{3}$, thus giving the solution
$\displaystyle y_n = c \left(\frac{1}{3}\right)^n + \frac{9}{2}$
Now use your initial condition to find c.
Note. If the difference equation is $\displaystyle y_{n+1} = \frac{1}{3 y_n} + 3$, it's harder problem.