We have the infinite series:

$\Sigma_{k=1}^{\infty} \frac{1}{7^{2k-1}} + \frac{2}{7^{2k}} = 9 \cdot \Sigma_{k=1}^{\infty} (\frac{1}{49})^{k} = 9 \cdot (\frac{1/49}{1-1/49}) = 9(\frac{1}{48}) = \boxed{\frac{3}{16}}.$

Using Turbo Pascal, we have the following code

# var a,b,c,i:real;i:longint;
# begin
# c:=0;
# for i:=1 to 22 do
# begin
# a:=1/(exp((2*i-1)*ln(7)));
# b:=1/(exp((2*i-1)*ln(7)));
# c:=a+b;
# end;
# write(c:0:10);
# readln;
# end.

After you press Crtl+F9, you'll see the result is $0.1875000000 \approx \frac{3}{16}$

Or you can do this

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

We can see this expression is an infinite geometric series with $d_{1}=\frac{1}{7}$ and $d_{2}=(\frac{1}{7})^{2}$ are common ratio. So that we have:

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