Let a =1+(^2 C_2)/(3!)+(^3 C_2)/(4!)+(^4 C_2)/(5!)+…; b =1+(^1 C_0+^1 C_1)/(1!)+(^2 C_0+^2 C_1+^2 C_2)/(2!)+(^3 C_0+^3 C_1+^3 C_2+^3 C_3)/(3!)+… Then…

Let $$\begin{aligned} & \mathrm{a}=1+\frac{{ }^2 \mathrm{C}_2}{3!}+\frac{{ }^3 \mathrm{C}_2}{4!}+\frac{{ }^4 \mathrm{C}_2}{5!}+\ldots \\ & \mathrm{b}=1+\frac{{ }^1 \mathrm{C}_0+{ }^1 \mathrm{C}_1}{1!}+\frac{{ }^2 \mathrm{C}_0+{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_2}{2!}+\frac{{ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_2+{ }^3 \mathrm{C}_3}{3!}+\ldots \end{aligned} $$ Then $\displaystyle \frac{2 b}{a^2}$ is equal to $\displaystyle \_\_\_\_$.