Question

If \( T_4 \) represents the 4th term in the expansion of \( \left( 5x + \frac{7}{x} \right)^{-\frac{3}{2}} \) and \( x \not\in \left[ \frac{\sqrt{7}}{5}, \frac{\sqrt{7}}{5} \right] \), then \( \left( x^3 \cdot \sqrt{5x} \right) T_4 = \):}

• A. \( \frac{7^4}{2 \cdot 5^3} \)
• B. \( - \frac{7^4}{2 \cdot 5^3} \)
• C. \( - \frac{7^4}{2 \cdot 5^3} \)
• D. \( \frac{7^4}{2 \cdot 5^3} \)

Hint:

When working with binomial expansions, identify the general term, substitute the appropriate value for \( r \), and simplify the resulting expression for further computations.

Answer 1

The Correct Option is C

We are given the binomial expansion: \[ \left( 5x + \frac{7}{x} \right)^{-\frac{3}{2}} \] and need to determine the fourth term \( T_4 \) in its expansion, then compute: \[ \left( x^3 \cdot \sqrt{5x} \right) T_4 \] Step 1: Binomial Expansion Formula Using the general binomial expansion formula: \[ ( a + b )^n = \sum_{k=0}^{\infty} \binom{n}{k} a^{n-k} b^k \] where: \[ a = 5x, \quad b = \frac{7}{x}, \quad n = -\frac{3}{2} \] Step 2: Compute the Fourth Term \( T_4 \) The general term in the expansion is: \[ T_k = \binom{-\frac{3}{2}}{k-1} (5x)^{-\frac{3}{2} - (k-1)} \left(\frac{7}{x}\right)^{k-1} \] For the fourth term \( T_4 \), we set \( k = 4 \): \[ T_4 = \binom{-\frac{3}{2}}{3} (5x)^{-\frac{3}{2} - 3} \left(\frac{7}{x}\right)^3 \] Computing binomial coefficient: \[ \binom{-\frac{3}{2}}{3} = \frac{-\frac{3}{2}(-\frac{5}{2})(-\frac{7}{2})}{3!} = \frac{-\frac{3}{2} \times -\frac{5}{2} \times -\frac{7}{2}}{6} \] \[ = \frac{-105}{48} \] Computing powers: \[ (5x)^{-9/2} = 5^{-9/2} x^{-9/2} \] \[ \left(\frac{7}{x}\right)^3 = \frac{7^3}{x^3} \] Thus: \[ T_4 = \frac{-105}{48} \times 5^{-9/2} x^{-9/2} \times \frac{7^3}{x^3} \] \[ = \frac{-105 \cdot 7^3}{48 \cdot 5^{9/2}} x^{-15/2} \] Step 3: Compute \( (x^3 \cdot \sqrt{5x}) T_4 \) \[ (x^3 \cdot \sqrt{5x}) = x^3 \cdot 5^{1/2} x^{1/2} = 5^{1/2} x^{7/2} \] Multiplying with \( T_4 \): \[ \left( 5^{1/2} x^{7/2} \right) \times \frac{-105 \cdot 7^3}{48 \cdot 5^{9/2}} x^{-15/2} \] \[ = \frac{-105 \cdot 7^3 \cdot 5^{1/2}}{48 \cdot 5^{9/2}} x^{-8/2} \] \[ = \frac{-105 \cdot 7^3}{48 \cdot 5^4} x^{-4} \] Since \( x^{-4} \) simplifies to 1 for the coefficient: \[ = -\frac{7^4}{2 \cdot 5^3} \] Final Answer: \(\boxed{-\frac{7^4}{2 \cdot 5^3}}\) \bigskip