Question

There is a question as following
\[0,\overline{1101}=\sum_{k=0}^{\infty}\frac{1}{2^{4k+1}}+\frac{1}{2^{4k+2}}+\frac{1}{2^{4k+4}}\] but i dont understand this following part where the nominators coming \[=\sum_{k=0}^{\infty}\frac{2^{2}.2^{4}+2.2^{4}+2.2^{2}}{2^{4k}.2^{7}}\]

Answer 1

i think you want to get common denominator

Answer 2

yes but even i know that too i couldnt recognize where is nominators coming

Answer 3

why they have this values...

Answer 4

\[\begin{align}0,\overline{1101}

&=\sum_{k=0}^{\infty}\frac{1}{2^{4k+1}}+\frac{1}{2^{4k+2}}+\frac{1}{2^{4k+4}}\\~\\
&=\sum_{k=0}^{\infty}\frac{2^3}{2^{4k+1}.2^3}+\frac{2^2}{2^{4k+2}.2^2}+\frac{1}{2^{4k+4}}\\~\\
&=\sum_{k=0}^{\infty}\frac{2^3}{2^{4k+4}}+\frac{2^2}{2^{4k+4}}+\frac{1}{2^{4k+4}}\\~\\

\end{align}\]

Answer 5

now that the denominators are same, you can add up the fractions

Answer 6

this is ok if it would be like that i would understand but it looks different

Answer 7

\[n^{(a+b)}=n^a~n^b\]
\[\frac{1}{2^{4k+1}}+\frac{1}{2^{4k+2}}+\frac{1}{2^{4k+4}}\]
\[\frac{1}{2^{4k}.2^1}+\frac{1}{2^{4k}.2^2}+\frac{1}{2^{4k}.2^4}\]
\[\frac{2^6}{2^{4k}.2^1.2^6}+\frac{2^5}{2^{4k}.2^2.2^5}+\frac{2^3}{2^{4k}.2^4.2^3}\]
\[\frac{2^6}{2^{4k}.2^7}+\frac{2^4}{2^{4k}.2^7}+\frac{2^3}{2^{4k}.2^7}\]
this is what it gets to me

Answer 8

the tops can factor to what is presented as well

Answer 9

why they choose that setup? i dont know

Answer 10

thank you both