(This supersedes an older post.) We have the BBP-type pi formulas,
Order 4.
$$\pi = \sum_{n=0}^\infty \frac{(-1)^n}{\color{blue}{4^{n}}}\left(\frac{2}{4n+1}+\frac{2}{4n+2}+\frac{1}{4n+3}\right)\tag1$$
Order 8.
$$\pi = \frac{1}{2}\sum_{n=0}^\infty \frac1{\color{blue}{4^{2n}}}\left(\frac{2^3}{8n+2}+\frac{4}{8n+3}+\frac{4}{8n+4}-\frac{1}{8n+7}\right)\tag2$$
Q: In general, for order $4m$ and positive integer $m$, is it true that, $$\large\pi = \frac{1}{4^{m-1}}\sum_{n=0}^\infty \frac{(-1)^{mn}}{4^{mn}}\Big(A_1+A_2+A_3\Big)$$
where,
$$A_1 = \sum_{k=0}^{m-1}(-1)^k\, \Big(\frac{2^{2m-2k-1}}{4mn+4k+1}\Big)$$
$$A_2 = \sum_{k=0}^{m-1}(-1)^k\, \Big(\frac{2^{2m-2k-1}}{4mn+4k+2}\Big)$$
$$A_3 = \sum_{k=0}^{m-1}(-1)^k\, \Big(\frac{2^{2m-2k-2}}{4mn+4k+3}\Big)$$
The case $m=7$ yielding order $28$ then answers the old post.