Define, $$\lambda_n =\frac{\tbinom{2n}{n}}{2^{2n}}=\frac{(\tfrac12)_n}{n!} $$

with binomial $\tbinom{n}{k}$ and Pochhammer symbol $(x)_n$. I noticed that the following 10 formulas have a nice "affinity".

Level 3:

$$\sum_{n=0}^\infty \lambda_n^3\, \frac{6n+1}{2^{2n}} =\frac{2^2}{\pi}\tag1$$ $$\sum_{\color{red}{n=1}}^\infty \frac1{n^3\lambda_n^3}\, \frac{6\big(n-\tfrac12\big)+1}{2^{2n}} =\pi^2\tag2$$

$$\sum_{n=0}^\infty \lambda_n^3\, \frac{42n+5}{2^{6n}} =\frac{2^4}{\pi}\tag3$$ $$\sum_{\color{red}{n=1}}^\infty \frac1{n^3\lambda_n^3}\, \frac{42\big(n-\tfrac12\big)+5}{2^{6n}} =\frac{\pi^2}3\tag4$$

Level 5:

$$\sum_{n=0}^\infty \lambda_n^5\, \frac{20n^2+8n+1}{(-2^2)^n}=\frac{2^3}{\pi^2}\tag5$$ $$\sum_{\color{red}{n=1}}^\infty \frac1{n^5\lambda_n^5}\, \frac{20\big(n-\tfrac12\big)^2+8\big(n-\tfrac12\big)+1}{(-2^2)^n} =-56\zeta(3)\tag6$$

$$\sum_{n=0}^\infty \lambda_n^5\, \frac{205n^2+45n+\tfrac{13}4}{(-2^{10})^n}=\frac{2^5}{\pi^2}\tag7$$ $$\sum_{\color{red}{n=1}}^\infty \frac1{n^5\lambda_n^5}\, \frac{205\big(n-\tfrac12\big)^2+45\big(n-\tfrac12\big)+\tfrac{13}4}{(-2^{10})^n} =-2\zeta(3)\tag8$$

Level 7:

$$\sum_{n=0}^\infty \lambda_n^7\, \frac{84n^3+38n^2+7n+\tfrac12}{2^{6n}} =\frac{2^4}{\pi^3}\tag9$$ $$\sum_{\color{red}{n=1}}^\infty \frac1{n^7\lambda_n^7}\, \frac{84\big(n-\tfrac12\big)^3+38\big(n-\tfrac12\big)^2+7\big(n-\tfrac12\big)+\tfrac12}{2^{6n}} =\frac{\pi^4}2\tag{10}$$

Most of these are scattered throughout the literature in various guises. See, for example, Guillera and Rogers' paper "Ramanujan Series Upside Down" which focuses only on level 3. The level 3 formulas for 1/pi were found by Ramanujan and can be explained by modular forms, while $(9)$ is by Gourevitch and $(10)$, in a different guise, is by MO user zy_. In this post, he remarked that Guillera, in private correspondence, considered it as new. (Consider that its partner was found by Gourevitch way back pre-2002.)

Q: However, what is the unifying theory for these ten formulas, and can we find paired examples for higher levels, like for $\zeta(5)$? (There is a Ramanujan-type formula for $\zeta(5)$ found by zy_ in the post cited, but it does not use $\lambda_n$ and doesn't seem to have a "partner".)