Alternating Sequence

$S_2 = S_1 + a_2 = (-1)^2a_2 + \frac{1}{2^2} \rightarrow S_1 = \frac{1}{4}$

$S_4 = S_3 + a_4 = (-1)^4a_4 + \frac{1}{2^4} \rightarrow S_3 = \frac{1}{16}$

$S_6 = S_5 + a_6 = (-1)^6a_6 + \frac{1}{2^6} \rightarrow S_5 = \frac{1}{64}$

$S_1 + S_3 + S_5 = \frac{1}{4} + \frac{1}{16} + \frac{1}{64} = \frac{21}{64}$ , so $a = 21$ , $b = 64$ , and $a + b = \boxed{85}$ .

$\begin{aligned} S_n & = (-1)^n a_n + \frac 1{2^n} & \small \color{#3D99F6} \text{Given} \\ \implies S_{n+1} & = (-1)^{n+1}a_{n+1} + \frac 1{2^{n+1}} & \small \color{#3D99F6} \text{For odd }n \\ S_{n_{\color{#3D99F6}odd}} + a_{n+1} & = a_{n+1} + \frac 1{2^{n+1}} \\ \implies S_{n_{\color{#3D99F6}odd}} & = \frac 1{2^{n+1}} \end{aligned}$

Therefore, $\begin{aligned} S_1 + S_3 + S_5 & = \frac 1{2^2} + \frac 1{2^4} + \frac 1{2^6} = \frac {2^4+2^2+1}{2^6} = \frac {21}{64} \end{aligned}$

$\implies a+b = 21+64 = \boxed{85}$ .

Following on from the answers given by observing $S_{2k} \implies S_{2k-1} = \frac{1}{2^{2k}}$ :

$S_{2k-1} = - a_{2k-1} + \frac{1}{2^{2k-1}} = \frac{1}{2^{2k}}$ .

Hence, $a_{2k-1} = \frac{1}{2^{2k-1}} - \frac{1}{2^{2k}} = \frac{1}{2^{2k}}$ .

---

Looking at $S_{2k-1} = (a_1 + a_2) + (a_3 + a_4) + ... + a_{2k-1} = \frac{1}{2^{2k}} = a_{2k-1}$ , and using a process of induction we get $a_{2k} = -a_{2k-1} = -\frac{1}{2^{2k}}$ .

So our sequence is: $\{ a_n \} = \frac{1}{4}, -\frac{1}{4}, \frac{1}{16}, -\frac{1}{16}, \frac{1}{64}, ...$