Question

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

• A. 10
• B. 8
• C. 15
• D. 12

Answer 1

The Correct Option is A

Step 1: General term in the binomial expansion

The general term in the expansion of \(\left(4x^5 - \frac{5}{2x}\right)^{2022}\) is given by:

\[ T_r = \binom{2022}{r-1} \cdot (4x^5)^{2022-(r-1)} \cdot \left(-\frac{5}{2x}\right)^{r-1} \]

Step 2: \(1011^{th}\) term from the beginning

The \(1011^{th}\) term from the beginning corresponds to \(r = 1011\):

\[ T_{1011} = \binom{2022}{1010} \cdot (4x^5)^{1012} \cdot \left(-\frac{5}{2x}\right)^{1010} \]

Step 3: \(1011^{th}\) term from the end

The \(1011^{th}\) term from the end corresponds to \(r = 1012\) from the beginning. Substituting \(r = 1012\):

\[ T_{1011}^{\text{end}} = \binom{2022}{1011} \cdot (4x^5)^{1010} \cdot \left(-\frac{5}{2x}\right)^{1012} \]

Step 4: Relationship between the two terms

According to the problem, the term from the end is 1024 times the term from the beginning:

\[ T_{1011}^{\text{end}} = 1024 \cdot T_{1011} \]

Substituting the expressions for the terms:

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

Step 5: Simplify the equation

Cancel common terms and simplify the powers of \(x\):

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

Simplifying further gives:

\[ \frac{1}{4x^2} = 1024 \]

Step 6: Solve for \(|x|\)

Rearrange the equation:

\[ x^2 = \frac{1}{4 \cdot 1024} = \frac{1}{4096} \]

Taking the square root:

\[ |x| = 10 \]

Final Answer

\(|x| = 10\)