The ratio of the coefficient of the middle te | vedclass

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(C) The middle term in the expansion of $(1+x)^{20}$ is the $\left(\frac{20}{2} + 1\right)^{th} = 11^{th}$ term.Its coefficient is $^{20}C_{10}$.The m

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$1$

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(C) The middle term in the expansion of $(1+x)^{20}$ is the $\left(\frac{20}{2} + 1\right)^{th} = 11^{th}$ term.Its coefficient is $^{20}C_{10}$.The middle terms in the expansion of $(1+x)^{19}$ are the $\left(\frac{19+1}{2}\right)^{th} = 10^{th}$ and $\left(\frac{19+1}{2} + 1\right)^{th} = 11^{th}$ terms.Their coefficients are $^{19}C_{9}$ and $^{19}C_{10}$.The sum of these coefficients is $^{19}C_{9} + ^{19}C_{10}$.Using the Pascal's identity $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$,we have $^{19}C_{9} + ^{19}C_{10} = ^{20}C_{10}$.Therefore,the required ratio is $\frac{^{20}C_{10}}{^{19}C_{9} + ^{19}C_{10}} = \frac{^{20}C_{10}}{^{20}C_{10}} = 1$.

The ratio of the coefficient of the middle term in the expansion of $(1+x)^{20}$ and the sum of the coefficients of two middle terms in the expansion of $(1+x)^{19}$ is $....$

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