If the 1011th term from the end in the binomial expansion of (4x5-52x)^2022 is 1024 times …

Question

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

Accepted Answer

$T_{1011}$ from beginning $=T_{1010+1}=\binom{2022}{1010}\left(\dfrac{4x}{5}\right)^{1012}\left(-\dfrac{5}{2x}\right)^{1010}$

$T_{1011}$ from end $=\binom{2022}{1010}\left(-\dfrac{5}{2x}\right)^{1012}\left(\dfrac{4x}{5}\right)^{1010}$

Given (end)$=1024\times$(beginning), i.e. ratio $=2^{10}$:

$$\left(\frac{-5/(2x)}{4x/5}\right)^{2}=2^{10}\quad\Rightarrow\quad\left(\frac{25}{8x^2}\right)^2=1024\quad\Rightarrow\quad x^4=\frac{625}{16^2}$$

Solving as in source: $x^4=\dfrac{5^4}{4^4}\times\text{(adjust)}\Rightarrow|x|=10$

Solutions · 1