If $A$ and $B$ are coefficients of ${x}^{n}$ in the expansions of ${ (1+x) }^{ 2n }$ and ${ (1+x) }^{ 2n-1 }$ respectively, then $\dfrac AB$ is equal to

We know that coefficient of ${x}^{r}$ in the expansion of ${ (1+x) }^{ m }$ is ${ _{ }^{ m }{ C } }_{ r }$

Thus $A={ _{ }^{ 2n }{ C } }_{ n }\quad ;\quad B={ _{ }^{ 2n-1 }{ C } }_{ n }$

We have

$\cfrac { A }{ B } =\cfrac { { _{ }^{ 2n }{ C } }_{ n } }{ { _{ }^{ 2n-1 }{ C } }_{ n } } =\cfrac { (2n)! }{ n!n! } \cfrac { (n!)(n-1)! }{ (2n-1)! } =\cfrac { 2n }{ n } =2$