If  $3^1\cdot {}^{20}{C}_0+\frac{3^2}{2}\cdot {}^{20}{C}_1+\frac{3^3}{3}\cdot {}^{20}{C}_2+\cdots$ up to 21 terms $=\frac{2^a-1}{b}$, then the value of $\left(\frac{{\displaystyle b}}{{\displaystyle a}}\right)$ is equal to

$3^1\cdot {}^{20}{C}_0+\frac{3^2}{2}\cdot {}^{20}{C}_1+\frac{3^3}{3}\cdot {}^{20}{C}_2+\cdots$up to 21 terms

$=\sum _{r=0}^{20}{3}^{r+1}\cdot \frac{{ }^{20}{C}_r}{r+1}=\sum _{r=0}^{20}{3}^{r+1}\cdot \frac{{ }^{21}{C}_{r+1}}{21}$

$=\frac{1}{21}\sum _{r=0}^{20}{3}^{r+1}\cdot {}^{21}{C}_{r+1}=\frac{1}{21}\left[{(1+3)}^{21}-1\right]$

$=\frac{4^{21}-1}{21}=\frac{2^{42}-1}{21}$

$\begin{array}{l}\Rightarrow a=42,b=21\\ \Rightarrow \frac{b}{a}=\frac{21}{42}=\frac{1}{2}\end{array}$