Restating the equation,

$x = \frac{1}{7^{1}} + \frac{2}{7^{2}} + \frac{1}{7^{3}} + \frac{2}{7^{4}} + ...$

We notice a pattern in the fractions; they can be separated like this:

$x = (\frac{1}{7^{1}} + \frac{1}{7^{3}} + ...) + (\frac{2}{7^{2}} + \frac{2}{7^{4}} + ...)$

Let $a = \frac{1}{7^{1}} + \frac{1}{7^{3}} + ...$ and let $b = \frac{2}{7^{2}} + \frac{2}{7^{4}} + ...$

Hence $x = a + b$

Lets solve for $a$ , $a = \frac{1}{7^{1}} + \frac{1}{7^{3}} + ...$

Multiplying both sides by $7^{2}$ we obtain,

$49a = 7 + \frac{1}{7^{1}} + \frac{1}{7^{3}} + ...$

Notice the RHS is equivalent to $7 + a$ ,

So $49a = 7 + a$

Results in: $\boxed{ a = \frac{7}{48} }$

And now, doing the same thing for b we obtain, $49b = 2 +b$

And hence, $\boxed{ b = \frac{2}{48} }$

And since $x = a + b$ ,

$\boxed{ x = \frac{7}{48} + \frac{2}{48} = \frac{9}{48} }$

$Let,\quad x=\frac { 1 }{ { 7 }^{ 1 } } +\frac { 2 }{ { 7 }^{ 2 } } +\frac { 1 }{ { 7 }^{ 3 } } +\frac { 2 }{ { 7 }^{ 4 } } +............\\ \Rightarrow 49x=7+2+\frac { 1 }{ { 7 }^{ 1 } } +\frac { 2 }{ { 7 }^{ 2 } } +\frac { 1 }{ { 7 }^{ 3 } } +\frac { 2 }{ { 7 }^{ 4 } } +............\\ \Rightarrow 49x=9+x\\ \Rightarrow x=\frac { 9 }{ 48 } =\frac { 3 }{ 16 }$

We can deal with the problem as the sum of two geometric sequences

1. $\frac { 1 }{ 7 } +\frac { 1 }{ 7^{ 3 } } +\frac { 1 }{ 7^{ 5 } } +...$ where ${ a}_{ n } =\frac { 1 }{ 7 }$ and $r =\frac {1 }{7^2 }$

2. $\frac { 2 }{ 7^2 } +\frac { 2 }{ 7^{ 4 } } +\frac { 2 }{ 7^{ 6 } } +...$ where ${ a}_{ n } =\frac { 2 }{ 7^2 }$ and $r =\frac {1 }{7^2 }$

Then we can get the sum of both of them using the infinite geometric series formula $\frac { { a }_{ n} }{ 1 - r }$ :

$\frac { \frac {1 }{7 } }{1- \frac { 1 }{ 7^2 } } +\frac { \frac { 2 }{7^2 } }{1 - \frac { 1 }{ 7^2 } }$ = $\boxed{\frac{3}{16}}$ which is the same as $\boxed{\frac{9}{48}}$

This can be expressed as the sum of two infinite geometric series, $\frac{1}{7^1}+\frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4} \cdots$ and $\frac{1}{7^2}+\frac{1}{7^4}+\frac{1}{7^6}+\frac{1}{7^8} \cdots$ . The sum of the first one is $\frac{\frac{1}{7}}{1-\frac{1}{7}}=\frac{1}{6}$ by the infinite geometric series formula, $\frac{a}{1-r}$ . The sum of the second one is $\frac{\frac{1}{49}}{1-\frac{1}{49}}=\frac{1}{48}$ . The total sum is $\frac{1}{6}+\frac{1}{48}=\boxed{\frac{9}{48}}$ .

add all the fraction given in the problem

(1/7+1/7^3+1/7^5+.......) +(2/7^2+2/7^4+2/7^6+............) 1/7(1+1/7^2+1/7^4+.........)+2/7^2(1+1/7^2+1/7^3+..........) (1+1/7^2+1/7^3+.........)(1/7+2/7^2) (1/1-(1/49))(1/7)(1+2/7) (49/48)(1/7)(9/7) 9/48