If the [1011^ th ] term from the end in the binominal expansion of [((4 x)/(5)-(5)/(2 x))^…

Question

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

Accepted Answer

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

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

$\mathrm{T}_{1011}$ from end

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

$$
\begin{aligned}
& \text { Given : }{ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1012}\left(\frac{4 \mathrm{x}}{5}\right)^{1010} \\
& =2^{10} \cdot{ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1010}\left(\frac{4 \mathrm{x}}{5}\right)^{1012}
\end{aligned}
$$

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

Solutions · 1