$ \begin{cases} V(k)=0 \text{ as } k < 1 \\ V(k+1) -V(k) = \left(\frac{1}{2}\right) \left(V(k) - V(k-1) \right) \text{ as } k \in [1,9] \\ V(k+1) = V(k) \text{ as } k >9 \end{cases} $
I have got
$V(10) - V(9) = (\frac{1}{2} (V(9) - V(8))$ so $1 = \left(\frac{3}{2}\right) V(9) - \frac{1}{2} V(8)$ and $V(9) = \frac{3}{2} V(8) - \frac{1}{2} V(7)$ ... $V(2) - V(1) = (\frac{1}{2} (V(1) - V(0))$ so $V(2) = \frac{3}{2} V(1)$
now I got
$V(3) = \frac{7}{6} V(2)$
n | V --------- 0 | 0 1 | V(1) 2 | 3/2 *V(1) 3 | 7/4 *V(1) 4 | 15/8 *V(1) 5 ... 6 7 8 9 .. 10 | 1
without calculating this by hands like this, is there some easy way to programmatically do this?