If A and B are coefficients of x^{n} in the e | vedclass

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(B) The coefficient of $x^{n}$ in the expansion of $(1+x)^{2n}$ is given by $A = {}^{2n}C_{n}$. The coefficient of $x^{n}$ in the expansion of $(1+x)^

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$2$

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(B) The coefficient of $x^{n}$ in the expansion of $(1+x)^{2n}$ is given by $A = {}^{2n}C_{n}$. The coefficient of $x^{n}$ in the expansion of $(1+x)^{2n-1}$ is given by $B = {}^{2n-1}C_{n}$. Now, we calculate the ratio $A / B$: $\frac{A}{B} = \frac{{}^{2n}C_{n}}{{}^{2n-1}C_{n}}$ Using the formula ${}^{n}C_{r} = \frac{n}{r} 	imes {}^{n-1}C_{r-1}$, we have ${}^{2n}C_{n} = \frac{2n}{n} 	imes {}^{2n-1}C_{n-1}$. However, a simpler approach is: $\frac{A}{B} = \frac{\frac{(2n)!}{n!n!}}{\frac{(2n-1)!}{n!(n-1)!}} = \frac{(2n)!}{n!n!} 	imes \frac{n!(n-1)!}{(2n-1)!} = \frac{2n 	imes (2n-1)!}{n 	imes (n-1)! 	imes n!} 	imes \frac{n!(n-1)!}{(2n-1)!} = \frac{2n}{n} = 2$.

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

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