(Re-upload) Infinite Sum With NO Infinite Sum

$\begin{aligned} \frac 27 + \frac 1{7^2} + \frac 2{7^3} + \frac 1{7^4} + \cdots & = \frac a9 + \frac b{9^2} + \frac a{9^3} + \frac b{9^4} + \cdots \\ \left(\frac 27 + \frac 1{7^2}\right) \left(1 + \frac 1{7^2} + \frac 1{7^4} + \cdots \right) & = \left(\frac a9 + \frac b{9^2}\right) \left(1 + \frac 1{9^2} + \frac 1{9^4} + \cdots \right) \\ \left(\frac {15}{49}\right) \left(\frac {49}{48} \right) & = \left(\frac {9a + b}{81}\right) \left(\frac {81}{80}\right) \\ \implies \frac {9a+b}{80} & = \frac 5{16} \\ 9a + b & = 25 \end{aligned}$

$\implies a = 2$ and $b=7$ for $a, b < 10$ . Therefore, $ab = \boxed {14}$ .

$\frac{2}{7}+\frac{1}{7^2}+\frac{2}{7^3}+\dots=\frac{2}{7}(1+\frac{1}{7^2}+\frac{1}{7^4}+\dots)+\frac{1}{7^2}(1+\frac{1}{7^2}+\frac{1}{7^4}+\dots)=(1+\frac{1}{7^2}+\frac{1}{7^4}+\dots)(\frac{2}{7}+\frac{1}{7^2})=\frac{1}{1-\frac{1}{7^2}}(\frac{2}{7}+\frac{1}{7^2})=\frac{5}{16}$

similar is done for the right hand side of the equation.

$\frac{a}{9}+\frac{}{}\frac{b}{9^2}\frac{a}{9^3}+\dots=\frac{1}{1-\frac{1}{9^2}}(\frac{a}{9}+\frac{b}{9^2})=\frac{9a+b}{80}$

therefore

$\frac{9a+b}{80}=\frac{5}{16} \implies 9a+b=25$

$9|25-b \ , \ 0<b<10 \implies b=7 \implies a=2 \implies a.b=14$

We can discuss the number system(ex. binary) under decimal places. As 102 means $1\times 10^2+2$ , 10.3 means $1\times 10+3\times 10^{-1}$ .

The left side of the equation can thus be rewritten as: $0.212121_{(7)}\cdots$ .

You would know the formula for making fraction form from circulating decimals: $\dfrac{cycle}{10^{length\ of\ cycle}-1}$

Now, we can extend this notation to general systems: $\left(\dfrac{cycle}{10^{length\ of\ cycle}-1}\right)_{(system)}$

Given $0.212121_{(7)}\cdots$ , it is equal to: $\left(\dfrac{21}{100-1}\right)_{(7)}=\dfrac{21}{66}_{(7)}=\dfrac{5}{16}$

Our goal is to make circulating 9-system decimals; take a few calculations to make new form: $\dfrac{5}{16}=\dfrac{25}{9^2-1}=\dfrac{27}{88}_{(9)}$

Hence, $\dfrac{27}{88}_{(9)}=0.272727_{(9)}\cdots \rightarrow a=2,\ b=7$ , having our answer $2\times7=\boxed{14}\$ .