# Summation of an Arithmetic, Parametric Sequence

I'm trying work on my ability to break complex patterns down,
and in this case I'm trying to model the denominators of Lacsap's Fractions:

I managed to get the sequence that represents the denominators:

But now I want to convert it to a series in order to get rid of the tedious term D_n-1.
I guess I'm just not quite sure how to apply the sum identity to a parametric sequence:

The red is the answer that I've found works, but I don't know how to get to it.
When I simply substitute k, D_1, and D_k with snippets from the sequence, I get it wrong.

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This is essentially OEIS A144216 – Henry Aug 16 '12 at 7:24

I think you wanted to write $$D_{n,r}=D_{n-1,r}+(n-r)$$ though you could have written $$D_{n,r}=D_{n-1,r-1}+r$$ with the combined effect that (assuming that you start from $D_{0,0}=0$) $$D_{n,r}=\sum_{i=1}^{n-r} i + \sum_{j=1}^{r} j = \frac{(n-r)(n-r+1)}{2} + \frac{r(r+1)}{2}$$ and hence your result.
Another way to look at it is to note that $D_{n,0}=D_{n,n}=\frac{n(n+1)}{2}$ and that $D_{n,r}=D_{n-r,0}+D_{r,r}$. For the numerator you have $N_{n,r}=\frac{n(n+1)}{2}$.
@Griffin: If you look at your triangle, each time you go down to the left you add the next integer $(1+2+\cdots)$ and every time you go down to the right you add the next integer $(1+2+\cdots)$ so the final figure is the sum of two triangle numbers. – Henry Aug 16 '12 at 11:58
$D(n,r) = D(n−1,r) +(n−r) = D(n−2,r) +(n-r-1) + (n−r) = \cdots$ $= D(r,r) + 1 +2 +\cdots + (n-r-1) + (n−r) = D(r,r) + \frac{(n-r)(n-r+1)}{2}$ while $D(r,r) = D(r-1,r-1) +r = D(r-2,r-2) +(r-1) +r = \cdots$ $= D(0,0) + 1 +2 +\cdots + (r-1) + r = D(0,0) + \frac{r(r+1)}{2}$. – Henry Aug 17 '12 at 0:33