Question

Convert the binary number (1.01010101...) × 2^3 to a decimal number g

Answer 1

`1.010101\ldots_2=1+\frac1{2^2}+\frac1{2^4}+\frac1{2^6}+\cdots`

Call the sum
`S`, and let's consider the
`n`-th partial sum denoted
`S_n`:


`S_n=1+\frac1{2^2}+\drac1{2^4}+\cdots+\frac1{2^(2n)}`

Now,


`\frac1{2^2}S_n=\frac1{2^2}+\frac1{2^4}+\frac1{2^6}+\cdots+\frac1{2^(2(n+1))}`

`\implies S_n-\frac1{2^2}S_n=1-\frac1{2^(2(n+1))}`

`\frac3{2^2}S_n=1-\frac1{2^(2(n+1))}`

Note that as
`n\to\infty`, the RHS approaches 1, so that


`S=\displaystyle\lim_(n\to\infty)S_n=\frac{2^2}3=\frac43`

Not sure what to make of the
`*2^3` part of your question, but perhaps you just need to multiply the above by 8?