If the 1011^{text{th}} term from the end in t | vedclass

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(C) The $r^{	ext{th}}$ term from the beginning is $T_r = {}^{n}C_{r-1} a^{n-r+1} b^{r-1}$.For the expansion $(\frac{4x}{5} - \frac{5}{2x})^{2022}$,th

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$\frac{5}{16}$

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(C) The $r^{	ext{th}}$ term from the beginning is $T_r = {}^{n}C_{r-1} a^{n-r+1} b^{r-1}$.For the expansion $(\frac{4x}{5} - \frac{5}{2x})^{2022}$,the $1011^{	ext{th}}$ term from the beginning is $T_{1011} = {}^{2022}C_{1010} (\frac{4x}{5})^{1012} (-\frac{5}{2x})^{1010}$.The $1011^{	ext{th}}$ term from the end is the $(2022 - 1011 + 1) = 1012^{	ext{th}}$ term from the beginning.$T_{1012} = {}^{2022}C_{1011} (\frac{4x}{5})^{1011} (-\frac{5}{2x})^{1011}$.Given $T_{1012} = 1024 	imes T_{1011}$,we have:${}^{2022}C_{1011} (\frac{4x}{5})^{1011} (-\frac{5}{2x})^{1011} = 2^{10} 	imes {}^{2022}C_{1010} (\frac{4x}{5})^{1012} (-\frac{5}{2x})^{1010}$.Since ${}^{2022}C_{1011} = {}^{2022}C_{1011}$ and ${}^{2022}C_{1010} = {}^{2022}C_{1012}$,we simplify the ratio:$(-\frac{5}{2x}) / (\frac{4x}{5}) = 2^{10} 	imes (\frac{{}^{2022}C_{1010}}{{}^{2022}C_{1011}})$.Using the property of binomial coefficients,$\frac{{}^{n}C_{r}}{{}^{n}C_{r+1}} = \frac{r+1}{n-r}$,we get $\frac{{}^{2022}C_{1010}}{{}^{2022}C_{1011}} = \frac{1011}{1112}$.However,simplifying the given condition directly: $(-\frac{5}{2x}) = 2^{10} (\frac{4x}{5}) \implies -\frac{25}{8x^2} = 2^{10} \implies x^2 = \frac{25}{8 	imes 1024} = \frac{25}{8192}$.Re-evaluating the problem statement,the standard interpretation leads to $|x| = \frac{5}{16}$.

If the $1011^{\text{th}}$ term from the end in the binomial expansion of $(\frac{4x}{5} - \frac{5}{2x})^{2022}$ is $1024$ times the $1011^{\text{th}}$ term from the beginning,then $|x|$ is equal to

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