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If $A$ and $B$ are coefficients of $x^n$ in the expansion of $(1+x)^{2 n}$ and $(1+x)^{2 n-1}$ respectively, then find $\frac{A}{B}$.
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Verified Answer
We have $(1+x)^{2 n}$
$$
\begin{aligned}
={ }^{2 n} C_0+{ }^{2 n} C_1 x+{ }^{2 n} C_2 x^2+\ldots .{ }^{2 n} C_n x^n+\ldots .+{ }^{2 n} C_{2 n} x^{2 n} \\
&(1+x) \text { (i) } \\
& \ldots .+{ }^{2 n-1} C_n x^n+\ldots .+{ }^{2 n-1}{ }^{2 n-1} C_{2 n-1} x^{2 n-1}
\end{aligned}
$$
According to the given data and equations (i) and (ii), we can claim that
$$
A={ }^{2 n} C_n \text { and } B={ }^{2 n-1} C_n
$$
$$
\Rightarrow \frac{A}{B}=\frac{{ }^{2 n} C_n}{{ }^{2 n-1} C_n}=\frac{\frac{2 n !}{n ! n !}}{\frac{(2 n-1) !}{n !(n-1) !}}
$$
$$
\Rightarrow \frac{A}{B}=\frac{2 n(2 n-1) !}{n(n-1) !} \times \frac{(n-1) !}{(2 n-1) !}=2 \Rightarrow A=2 B
$$
$$
\begin{aligned}
={ }^{2 n} C_0+{ }^{2 n} C_1 x+{ }^{2 n} C_2 x^2+\ldots .{ }^{2 n} C_n x^n+\ldots .+{ }^{2 n} C_{2 n} x^{2 n} \\
&(1+x) \text { (i) } \\
& \ldots .+{ }^{2 n-1} C_n x^n+\ldots .+{ }^{2 n-1}{ }^{2 n-1} C_{2 n-1} x^{2 n-1}
\end{aligned}
$$
According to the given data and equations (i) and (ii), we can claim that
$$
A={ }^{2 n} C_n \text { and } B={ }^{2 n-1} C_n
$$
$$
\Rightarrow \frac{A}{B}=\frac{{ }^{2 n} C_n}{{ }^{2 n-1} C_n}=\frac{\frac{2 n !}{n ! n !}}{\frac{(2 n-1) !}{n !(n-1) !}}
$$
$$
\Rightarrow \frac{A}{B}=\frac{2 n(2 n-1) !}{n(n-1) !} \times \frac{(n-1) !}{(2 n-1) !}=2 \Rightarrow A=2 B
$$
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