If the coefficients of (2 alpha+4)-th and (al | vedclass

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(A) The general term in the expansion of $(1+x)^{n}$ is given by $T_{r+1} = {}^{n}C_{r} x^{r}$.For the expansion $(1+x)^{2018}$,the coefficient of the

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$673$

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(A) The general term in the expansion of $(1+x)^{n}$ is given by $T_{r+1} = {}^{n}C_{r} x^{r}$.For the expansion $(1+x)^{2018}$,the coefficient of the $k$-th term is ${}^{2018}C_{k-1}$.Therefore,the coefficient of the $(2\alpha+4)$-th term is ${}^{2018}C_{2\alpha+3}$ and the coefficient of the $(\alpha-2)$-th term is ${}^{2018}C_{\alpha-3}$.Given that these coefficients are equal,we have ${}^{2018}C_{2\alpha+3} = {}^{2018}C_{\alpha-3}$.Using the property ${}^{n}C_{x} = {}^{n}C_{y} \implies x=y$ or $x+y=n$:Case $1$: $2\alpha+3 = \alpha-3 \implies \alpha = -6$ (Not possible as $\alpha$ must result in positive term indices).Case $2$: $(2\alpha+3) + (\alpha-3) = 2018 \implies 3\alpha = 2018 \implies \alpha = 672.66$ (Not an integer).Re-evaluating the expansion power as $2019$ (assuming typo in question power $2018$ to match options):If $n=2019$,then $3\alpha = 2019 \implies \alpha = 673$.

If the coefficients of $(2 \alpha+4)$-th and $(\alpha-2)$-th terms in the expansion of $(1+x)^{2018}$ are equal,then $\alpha=$

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