I have the following recursive relation ($k$ and $j$ natural numbers):
$$ r_k = r_{k+1} + 2(k+1) $$ And for a specific $j$ $$r_j = 0$$
How can I find $r_{k,j}$?
I know the solution, that is:
$$r_{k,j} = j(j+1) - k(k+1)$$
I tried by handling the finite differences, namely:
$$ \frac{\Delta r}{\Delta k} = \frac{r_{k+1} -r_k}{(k+1)-k} = r_{k+1} -r_k =-2(k+1) $$
And then I integrated $\frac{dr}{dk}$ with the condition $r(j)=0$.
But I did not find the expected result.
Note that $$\begin{align}r_n&=\sum_{k=j}^{n-1}(r_{k+1}-r_{k})+r_j=\sum_{k=j}^{n-1}(-2(k+1))+0\\ &=-2\sum_{k=0}^{n-1}(k+1)+2\sum_{k=0}^{j-1}(k+1) =-2\sum_{k=1}^{n}k+2\sum_{k=1}^{j}k\\ &=-n(n+1)+j(j+1). \end{align}$$