$\frac{1}{2}$ + $\frac{1}{3}$- $\frac{1}{4}$ + $\frac{1}{8}$ + $\frac{1}{9}$ - $\frac{1}{16}$ + $\frac{1}{32}$ +$\frac{1}{27}$ - $\frac{1}{64}$....
$\Rightarrow$ $\frac{{\color{Red} 1}}{{\color{Red} 2}}$ + $\frac{{\color{Green} 1}}{{\color{Green} 3}}$ - $\frac{1}{4}$ + $\frac{{\color{Red} 1}}{{\color{Red} 8}}$ + $\frac{{\color{Green} 1}}{{\color{Green} 9}}$ - $\frac{1}{16}$+ $\frac{{\color{Red} 1}}{{\color{Red} 3 \color{Red}2}}$ + $\frac{{\color{Green} 1}}{{\color{Green} 2\color{Green} 7}}$ - $\frac{1}{64}$.....
$\Rightarrow$ $\frac{{\color{Red} 1}}{{\color{Red} 2}}$ + $\frac{{\color{Red} 1}}{{\color{Red} 8}}$ + $\frac{{\color{Red} 1}}{{\color{Red} 3 \color{Red}2}}$ +... +$\frac{{\color{Green} 1}}{{\color{Green} 3}}$+ $\frac{{\color{Green} 1}}{{\color{Green} 9}}$+ $\frac{{\color{Green} 1}}{{\color{Green} 2\color{Green} 7}}$+...- $\frac{1}{4}$- $\frac{1}{16}$- $\frac{1}{64}$-...
$\Rightarrow$ $\left ( \frac{{\color{Red} 1}}{{\color{Red} {2^{1}}}}+ \frac{{\color{Red} 1}}{{\color{Red}{ 2^{3}}}} + \frac{{\color{Red} 1}}{\color{Red}{2^{5}}}+...\right )$ + $\left ( \frac{{\color{Green} 1}}{{\color{Green} 3}}+ \frac{{\color{Green} 1}}{{\color{Green} { 3^{2}}}}+ \frac{{\color{Green} 1}}{{\color{Green}{ 3^{3}} }} +... \right )$-$\left (\frac{1}{2^{2}}+ \frac{1}{2^{4}}+\frac{1}{2^{6}}+... \right )$
Hence the answer is the sum of 3 infinite G.P. series.
Sum of infinite G.P. Series = $\frac{a}{1-r}$
$\left ( \frac{{\color{Red} 1}}{{\color{Red} {2^{1}}}}+ \frac{{\color{Red} 1}}{{\color{Red}{ 2^{3}}}} + \frac{{\color{Red} 1}}{\color{Red}{2^{5}}}+...\right )$ = $\frac{{\color{Red} 1}}{{\color{Red} 2}}\left ( {\color{Red} 1} + \frac{{\color{Red} 1}}{{\color{Red} {2^{2}}}} + \frac{{\color{Red} 1}}{{\color{Red} {2^{4}}}} + ...\right )$ = $\frac{{\color{Red} 1}}{{\color{Red} 2}}\left ( \frac{\color{Red}1}{{\color{Red} {1-\frac{1}{4}}}} \right )$ =$\frac{{\color{Red} 1}}{{\color{Red} 2}} * \frac{{\color{Red} 4}}{{\color{Red} 3}}$ = $\frac{{\color{Red} 2}}{{\color{Red} 3}}$
$\left ( \frac{{\color{Green} 1}}{{\color{Green} 3}}+ \frac{{\color{Green} 1}}{{\color{Green} { 3^{2}}}}+ \frac{{\color{Green} 1}}{{\color{Green}{ 3^{3}} }}+... \right )$ = $\frac{{\color{Green} 1}}{{\color{Green} 3}} \left ( {\color{Green} 1}+ \frac{{\color{Green} 1}}{{\color{Green} 3}} + \frac{{\color{Green} 1}}{{\color{Green} {3^{2}}}} +...\right )$ = $\frac{{\color{Green} 1}}{{\color{Green} 3}} \left ( \frac{{\color{Green} 1}}{{\color{Green} {1 - \frac{1}{3} }}} \right )$ = $\frac{{\color{Green} 1}}{{\color{Green} 3}} * \frac{{\color{Green} 3}}{{\color{Green} 2}}$ = $\frac{{\color{Green} 1}}{{\color{Green} 2}}$
$\left (\frac{1}{2^{2}}+ \frac{1}{2^{4}}+\frac{1}{2^{6}}+... \right )$ = $\frac{1}{4} \left (1+ \frac{1}{2^{2}}+\frac{1}{2^{4}}+... \right )$ = $\frac{1}{4}\left ( \frac{1}{1-\frac{1}{4}} \right )$ = $\frac{1}{4} * \frac{4}{3}$ = $\frac{1}{3}$
$\frac{{\color{Red} 2}}{{\color{Red} 3}} + \frac{{\color{Green} 1}}{{\color{Green} 2}} - \frac{1}{3}$ = $\frac{{\color{Red} 4 }+ {\color{Green} 3} -2}{6}$ = $\frac{5}{6}$
$\therefore$ Option B $\frac{5}{6}$ is the right answer.