[tex]1.010101\ldots_2=1+\dfrac1{2^2}+\dfrac1{2^4}+\dfrac1{2^6}+\cdots[/tex]
Call the sum [tex]S[/tex], and let's consider the [tex]n[/tex]-th partial sum denoted [tex]S_n[/tex]:
[tex]S_n=1+\dfrac1{2^2}+\drac1{2^4}+\cdots+\dfrac1{2^{2n}}[/tex]
Now,
[tex]\dfrac1{2^2}S_n=\dfrac1{2^2}+\dfrac1{2^4}+\dfrac1{2^6}+\cdots+\dfrac1{2^{2(n+1)}}[/tex]
[tex]\implies S_n-\dfrac1{2^2}S_n=1-\dfrac1{2^{2(n+1)}}[/tex]
[tex]\dfrac3{2^2}S_n=1-\dfrac1{2^{2(n+1)}}[/tex]
Note that as [tex]n\to\infty[/tex], the RHS approaches 1, so that
[tex]S=\displaystyle\lim_{n\to\infty}S_n=\frac{2^2}3=\frac43[/tex]
Not sure what to make of the [tex]\times2^3[/tex] part of your question, but perhaps you just need to multiply the above by 8?
