Suppose A and B are the coefficients of 30^{t | vedclass

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(B) The $r^{	ext{th}}$ term in the expansion of $(1+x)^m$ is given by $T_r = {}^{m}C_{r-1} x^{r-1}$.For the expansion of $(1+x)^{2n-1}$:The $30^{	ex

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$21$

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(B) The $r^{	ext{th}}$ term in the expansion of $(1+x)^m$ is given by $T_r = {}^{m}C_{r-1} x^{r-1}$.For the expansion of $(1+x)^{2n-1}$:The $30^{	ext{th}}$ term is $T_{30} = {}^{2n-1}C_{29} x^{29}$,so $A = {}^{2n-1}C_{29}$.The $12^{	ext{th}}$ term is $T_{12} = {}^{2n-1}C_{11} x^{11}$,so $B = {}^{2n-1}C_{11}$.Given $2A = 5B$,we have $2({}^{2n-1}C_{29}) = 5({}^{2n-1}C_{11})$.Using the property ${}^{n}C_r = {}^{n}C_{n-r}$,we know ${}^{2n-1}C_{29} = {}^{2n-1}C_{(2n-1)-29} = {}^{2n-1}C_{2n-30}$.Thus,$2({}^{2n-1}C_{2n-30}) = 5({}^{2n-1}C_{11})$.Using the ratio formula $\frac{{}^{n}C_r}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$,we can solve for $n$ by comparing the terms or testing the options.For $n=21$,$2n-1 = 41$.$2({}^{41}C_{29}) = 5({}^{41}C_{11})$.Since ${}^{41}C_{29} = {}^{41}C_{12}$,we check $2({}^{41}C_{12}) = 5({}^{41}C_{11})$.$2 	imes \frac{41!}{12! 29!} = 5 	imes \frac{41!}{11! 30!}$.$2 	imes \frac{1}{12} = 5 	imes \frac{1}{30} \implies \frac{2}{12} = \frac{5}{30} \implies \frac{1}{6} = \frac{1}{6}$.Thus,$n = 21$ is the correct answer.

Suppose $A$ and $B$ are the coefficients of $30^{\text{th}}$ and $12^{\text{th}}$ terms respectively in the binomial expansion of $(1+x)^{2n-1}$. If $2A = 5B$,then $n$ is equal to:

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