Solving: $\sum\limits_{k=3}^{n}\bigg\lfloor | \frac {-1-\sqrt {1+8k}} {2}|\bigg\rfloor$

Question

How to do this:

$$\sum\limits_{k=3}^{n}\bigg\lfloor | \frac {-1-\sqrt {1+8k}} {2}|\bigg\rfloor$$ Where $\lfloor.\rfloor$ denotes floor function and $|.|$ denotes absolute value.

Any hints, method, would help me a lot.

Answer 1

First of all, observe (trivially) that the sum can be rewritten as:

$$\sum\limits_{k=3}^{n}\bigg\lfloor \frac {1+\sqrt {1+8k}} {2}\bigg\rfloor$$

Secondly, @dxiv's comment is a very good hint.

To see why that's so, consider the formula for the $(n-1)$th triangular number $k$. This is given by $\frac 12 (n)(n-1) = k$.

Solving for $n$ using the quadratic formula gives:

$\displaystyle n = \frac{1 + \sqrt{1 + 8k}}{2}$, which is only integral when $k$ is a triangular number as mentioned in the comment.

All that's left is to observe the pattern in the series. It starts:

$3,3,3,4,4,4,4,5,5,5,5,5,6...$

where it should be obvious that there are $m$ occurences of the natural number $m$ for $m \geq 3$.

I think you should be able to proceed from here.