Prove $\sum_{i=1}^n{(-1)^{i(i-1)/2}}$ is bounded

Question

I tried to prove this using Mathematical Induction (omitting base step):

Inductive step:

If ${\sum_{i=1}^n{(-1)^{i(i-1)/2 } \le 1}}$ then ${\sum_{i=1}^{n+1}{(-1)^{i(i-1)/2 } \le 1}}$

But at this point, I got stuck when arguing what happens when (n+1)n/2 is even. So I tried a different aproach: I realized that if you take 4 consecutive terms and add them, you get zero no matter how you choose those terms. This, plus making use of complete induction, could let me say that:

${\sum_{i=1}^{n+1}{(-1)^{i(i-1)/2 } = {\sum_{i=1}^{n-3}{(-1)^{i(i-1)/2 }\le1} } }}$

However I did not found a way to prove this, and maybe I am just going the wrong way with this making it harder than what it should be.

Any help/hint would be really appreciated. Thank you for your time.

Answer 1

Since

$$\frac{i(i-1)}2=\sum_{n=1}^{i-1}n=1+2+3+\dots(n-1)$$

It is easy to see that for $i>1$, every $4$ terms will have $2$ even and $2$ odd terms. Since $(-1)^n$ will cancel with itself every four terms, the problem reduces by induction down to the bounds of

$$\sum_{i=1}^{1,2,3,4}(-1)^{i(i-1)/2}$$

which can be easily calculated.