I need to evaluate this series.Without using derivative.
$$A=\frac{2} {2^2} + \frac{4} {2^5}+ \frac{6} {2^8} + \cdots $$
Where the $i$ th member is calculated with the the formula below:
$$A_i=\frac{i} {2^{3i-2}}$$
Feel free to edit the tags please.Thanks for you help.
$$a=\frac{2}{2^2}+\frac{4}{2^5}+\frac{6}{2^8}+\frac{8}{2^11}...+ $$multiply a by $$\frac{1}{2^3} $$ then find $$a -\frac{1}{2^3}a $$ $$a -\frac{1}{2^3}a=\\\frac{2}{2^2}+\frac{4}{2^5}+\frac{6}{2^8}+\frac{8}{2^{11}}...\\-(\frac{2}{2^5}+\frac{4}{2^8}+\frac{6}{2^11}+\frac{8}{2^{14}}...)\\a -\frac{1}{2^3}a=\frac{2}{2^2}+\frac{4-2}{2^5}+\frac{6-4}{2^8}+\frac{8-6}{2^{11}}...$$now you have $$\frac{7}{8}a=2(\frac{1}{2^2}+\frac{1}{2^5}+\frac{1}{2^8}+\frac{1}{2^{11}}...)$$
