recurrence relations - closed form for $d(4)=2$, $d(n+1)=d(n)+n-1$?

I am helping a friend in his last year of high school with his math class. They are studying recurrences and proof by inference. One of the exercises was simply "How many diagonals does a regular $n$-polygon have?".

We quickly found a direct answer to that question: $d(n) = \frac{(n-3)n}2$, but since they are studying recurrences, we didn't stop, and found the recurrence $d(n+1)=d(n)+n-1$. We proceeded to prove by induction that the closed form we initially found is indeed a solution for the recurrence.

But then he asked me a question I couldn't answer: we found the closed form because the question gives us an obvious mapping to an easy to solve geometry problem, but is there a way to find a closed form using the recurrence only?

Answer

Yes it is possible. This is an example of a telescoping sequence.

Notice that $d_{n+1} - d_n = n-1$

Now consider

$(d_{n+1} - d_n) + (d_n - d_{n-1}) + (d_{n-1} - d_{n-2}) + \cdots + (d_5 - d_4)$

I will leave the rest to you.