If the 1011^ term from the end in the binominal expansion of ((4x)/(5)-(5)/(2x))^2022 is 1…

Question

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

Accepted Answer

${T}_{1011}$ from beginning $={T}_{1010+1}$

$={}^{2022}{C}_{1010}{(\frac{4x}{5})}^{1012}{(\frac{-5}{2x})}^{1010}$

${T}_{1011}$ from end

$={}^{2022}{C}_{1010}{(\frac{-5}{2x})}^{1012}{(\frac{4x}{5})}^{1010}$

$\begin{gathered} {}^{2022}{C}_{1010}{(\frac{-5}{2x})}^{1012}{(\frac{4x}{5})}^{1010} \ ={2}^{10}\cdot {}^{2022}{C}_{1010}{(\frac{-5}{2x})}^{1010}{(\frac{4x}{5})}^{1012} \end{gathered}$

$\begin{gathered} \Rightarrow {(\frac{-5}{2x})}^{2}={2}^{10}{(\frac{4x}{5})}^{2} \ \Rightarrow {x}^{4}=\frac{{5}^{4}}{{2}^{16}} \ \Rightarrow |x|=\frac{5}{16} \end{gathered}$

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