What is the Solution to this sum $\sum \limits_{n=1}^{\infty}(1-(-1)^{\frac{n(n+1)}{2}})(\frac{1}{2})^n$ what is the value of this series $\sum \limits_{n=1}^{\infty}(1-(-1)^{\frac{n(n+1)}{2}})(\frac{1}{2})^n$   ?
Anything what's solid and that i got so far is only
$\sum \limits_{n=1}^{\infty}(1-(-1)^{\frac{n(n+1)}{2}})(\frac{1}{2})^n$ = $\sum \limits_{n=1}^{\infty}(1-i^{n^2+n})(\frac{1}{2})^n$
and that the right part reminds me of the geometric series
 A: Note that$$(1-(-1)^{\frac{n(n+1)}{2}})=\begin{cases}0 & \text{ if } n \equiv 0,3 \pmod{4} \\ 2 & \text{ if } n \equiv 1,2 \pmod{4}\end{cases}$$
Thus
\begin{align*}
\sum_{n=1}^{\infty}(1-(-1)^{\frac{n(n+1)}{2}})\left(\frac{1}{2}\right)^n&=\sum_{\substack{n=1 \\{\small n \equiv 0 \pmod{4}}}}^{\infty}0+\sum_{\substack{n=1 \\{\small n \equiv 1 \pmod{4}}}}^{\infty}+\sum_{\substack{n=1 \\{\small n \equiv 2 \pmod{4}}}}^{\infty}+\sum_{\substack{n=1 \\{\small n \equiv 3 \pmod{4}}}}^{\infty}0\\
&= \sum_{\substack{n=1 \\{\small n \equiv 1 \pmod{4}}}}^{\infty}2 \left(\frac{1}{2}\right)^n+\sum_{\substack{n=1 \\{\small n \equiv 2 \pmod{4}}}}^{\infty}2 \left(\frac{1}{2}\right)^n\\
&= 2\left(\sum_{\substack{n=1 \\{\small n \equiv 1 \pmod{4}}}}^{\infty}\frac{1}{2^n}+\sum_{\substack{n=1 \\{\small n \equiv 2 \pmod{4}}}}^{\infty} \frac{1}{2^n}\right)\\
\end{align*}
Now we just have to focus on these geometric series. For example,
\begin{align*}
\sum_{\substack{n=1 \\{\small n \equiv 1 \pmod{4}}}}^{\infty}\frac{1}{2^n}&=\frac{1}{2}+\frac{1}{2^5}+\frac{1}{2^9}+\dotsb\\
&=\frac{\frac{1}{2}}{1-\frac{1}{2^4}}\\
&=\frac{8}{15}.
\end{align*}
Hopefully you can complete now.
