Find the radius of convergence of the power series $\displaystyle\sum_{n=0}^{\infty}a_nz^n$, where $a_{2k+1} = 2^k$ and $a_{2k} = (1 + (1/k))^2$ for $k = 0, 1, 2, \dotsc$

I started off by doing the ratio test, but I know that the ration test is for $|a_{k+1} |/|a_k|$ and here they throw in the $2k+1$ so I am utterly lost and don't know how to start. can anyone help?
Cauchy-Hadamard makes it really simple here since

$$\lim_{n\to\infty}\sup\sqrt[n]{a_n}=\lim_{n\to\infty}\sup\begin{cases}\sqrt[n]{2^n}\\{}\\\sqrt[n]{\left(1+\frac1n\right)^2}\end{cases}\;\;=2$$

and the radius of convergence is thus $\;1/2\;$ .

Using the ratio test for odd/even-indexed subsequences:

$$\frac{a_{2n+2}}{a_{2n}}=\frac{\left(1+\frac1{n+1}\right)^2}{\left(1+\frac1n\right)2}\xrightarrow[n\to\infty]{}1$$

$$\frac{a_{2n+1}}{a_{2n-1}}=\frac{2^n}{2^{n-1}}=2$$

and we're done (why? Why can we split the sequence as above and do the ratio test separatedly) and then we choose the minimal one?)
Cauchy-Hadamard makes it really simple here since

$$\lim_{n\to\infty}\sup\sqrt[n]{a_n}=\lim_{n\to\infty}\sup\begin{cases}\sqrt[n]{2^n}\\{}\\\sqrt[n]{\left(1+\frac1n\right)^2}\end{cases}\;\;=2$$

and the radius of convergence is thus $\;1/2\;$ .

Using the ratio test for odd/even-indexed subsequences:

$$\frac{a_{2n+2}}{a_{2n}}=\frac{\left(1+\frac1{n+1}\right)^2}{\left(1+\frac1n\right)2}\xrightarrow[n\to\infty]{}1$$

$$\frac{a_{2n+1}}{a_{2n-1}}=\frac{2^n}{2^{n-1}}=2$$

and we're done (why? Why can we split the sequence as above and do the ratio test separatedly) and then we choose the minimal one?)