Summations and Infinite Series, oh my!

$S = \frac{5}{16} + \frac{5^{2}}{16^{2}} + \frac{5^{3}}{16^{3}} ............$

$\frac{S}{16} = \frac{5}{16^{2}} + \frac{5^{2}}{16^{3}} + ..............$

Subtracting both,

$S - \frac{S}{16} = \frac{5}{16} + \frac{5.4}{16^{2}} + \frac{5^{2}.4}{16^{3}} +....$

$\frac{15S}{16} = \frac{5}{16} + \frac{5.4}{16^{2}}(1 + \frac{5}{16} + \frac{5^{2}}{16^{2}} + ....$

$S = \frac{5}{11}$

$4(\frac{\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} ........}{44})$

$= \frac{1}{22}$

Adding both,

$= \frac{1}{2}$

$\displaystyle \sum_{n=0}^{4} 1^{n} + 2^{n} = 1 + 1 + 1 + 1 + 1 + 2^{0} + 2 + 2^{2} + .. + 2^{4}$

$= 5 + \frac{2^{5} - 1}{2-1} = 36$

So, here's my first attempt at the no-word proofs. Could anybody please rate this proof/ tell me how I can improve it?

$S=\frac{5}{16}+\frac{25}{256}+\frac{125}{4096}$ .

$\frac{16}{5}S=\frac{16}{5}(\frac{5}{16}+\frac{25}{256}+\frac{125}{4096})=1+S$ .

$S=\frac{5}{11}$

$y=\frac{4}{3}+\frac{4}{9}+\frac{4}{27}+...$

$3y=3(\frac{4}{3}+\frac{4}{9}+\frac{4}{27}+...)=4+y$

$y=2$ .

$\frac{2}{44}=\frac{1}{22}$ .

$S+\frac{y}{44}=\frac{5}{11}+\frac{1}{22}=\frac{10}{22}+\frac{1}{22}=\frac{11}{22}=\frac{1}{2}$

$1+1+1+2+1+4+1+8+1+16=36$

Use a/1-r for the two infinite geometric series. For the first one a = 5/16 and r = 5/16. For the second one a = 1/33 and r = 1/3. Therefore, the first series is equal to 5/11 and the second series is equal to 1/22 and a/b = 1/2. By calculation the summation, the answer 26 is achieved.