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

Question

Let \( \alpha, \beta, \gamma \) and \( \delta \) be the coefficients of \( x^7, x^5, x^3, x \) respectively in the expansion of \( (x + \sqrt{x^3 - 1})^5 + (x - \sqrt{x^3 - 1})^5, \, x > 1 \). If \( \alpha u + \beta v = 18 \), \( \gamma u + \delta v = 20 \), then \( u + v \) equals:

Answer 1

To solve this problem, we need to understand the concept of binomial expansion and how terms are counted from the start and the end.

The general term for the binomial expansion of (a + b)^n is given by:

T_{r+1} = \binom{n}{r} a^{n-r} b^r

Here, we need to consider the binomial expansion of \left(\frac{4x}{5} - \frac{5}{2x}\right)^{2022}.

Identify the terms:

a = \frac{4x}{5},
b = -\frac{5}{2x},
n = 2022.

From the end, the {1011}^{th} term is equivalent to the {1012}^{th} term from the beginning (since there are 2023 terms in total for (n+1 = 2023)).

Using the general term formula:

The {1011}^{th} term from the beginning can be represented by T_{1011} = \binom{2022}{1010} \left(\frac{4x}{5}\right)^{1012} \left(-\frac{5}{2x}\right)^{1010}.
The {1011}^{th} term from the end is T_{1012} = \binom{2022}{1011} \left(\frac{4x}{5}\right)^{1011} \left(-\frac{5}{2x}\right)^{1011}.

Given that the {1011}^{th} term from the end is 1024 times the {1011}^{th} term from the start, we have:

|\binom{2022}{1011} \times \left(\frac{4x}{5}\right)^{1011} \times \left(-\frac{5}{2x}\right)^{1011}| = 1024 \times |\binom{2022}{1010} \times \left(\frac{4x}{5}\right)^{1012} \times \left(-\frac{5}{2x}\right)^{1010}|

Simplifying, we get:

\left|\frac{1}{4x/5}\right| = 1024

Hence:

\left|\frac{5}{4x}\right| = 1024

Solving gives:

x = \pm 10

So, |x| = 10.

Therefore, the correct answer is 10.

Answer 2