Question

If the constant term in the expansion of $\left(\frac{\sqrt[5]{3}}{x}+\frac{2x}{\sqrt[3]{5}}\right)^{12}$, $x \neq 0$, is $\alpha \times 2^8 \times \sqrt[5]{3}$, then $25\alpha$ is

• A. 639
• B. 724
• C. 693
• D. 742

Answer 1

The Correct Option is C

\[ T_{r+1} = \binom{12}{r} \left(\frac{3^{1/5}}{x}\right)^{12-r} \left(\frac{2x}{5^{1/3}}\right)^r \]

\[ T_{r+1} = \binom{12}{r} \frac{3^{12-r}/5}{x^{12-r}} \frac{(2x)^r}{(5)^{r/3}} \]

Given \( r = 6 \), \[ T_7 = \binom{12}{6} \frac{(3^{6/5})(2^6)(x^6)}{x^6 (5^2)} = \binom{12}{6} \frac{3^6 \cdot 2^6}{5^2} \]

Simplifying further, \[ T_7 = \frac{9 \cdot 11 \cdot 7}{25} \cdot 2^8 \cdot 3^{1/5} \]

Finally, equating, \[ 25\alpha = 693 \]

Final Answer: 693

Answer 2

Step 1: General term in the expansion
Consider the term from the expansion of \((\frac{\sqrt[5]{3}}{x} + \frac{2x}{\sqrt[3]{5}})^{12}\):
T_r+1 = ¹²Cᵣ × \((\frac{\sqrt[5]{3}}{x})^{12−r}\) × \((\frac{2x}{\sqrt[3]{5}})^r\).

Step 2: Simplify T_r+1
T_r+1 = ¹²Cᵣ × (\( \sqrt[5]{3} \))^12−r × (2)^r × x^r−(12−r) × (5)^−r/3.
⟹ T_r+1 = ¹²Cᵣ × 2ʳ × 3^(12−r)/5 × 5^−r/3 × x^(2r−12).

Step 3: For constant term
Power of x must be 0.
So, 2r − 12 = 0 ⟹ r = 6.

Step 4: Substitute r = 6
T₇ = ¹²C₆ × 2⁶ × 3^(12−6)/5 × 5^−6/3.
Simplify: T₇ = ¹²C₆ × 2⁶ × 3^6/5 × 5⁻².
= (¹²C₆ / 25) × 2⁶ × 3^6/5.

Step 5: Express in given form
Given that constant term = α × 2⁸ × ³√[5]{3}.
So, equate:
α × 2⁸ × 3^1/5 = (¹²C₆ / 25) × 2⁶ × 3^6/5.

Simplify:
α = [(¹²C₆ / 25) × 2⁶ × 3^6/5] / [2⁸ × 3^1/5]
= (¹²C₆ / 25) × (1 / 2²) × 3.
= (¹²C₆ × 3) / (25 × 4).

Step 6: Compute numerical value
¹²C₆ = 924.
So, α = (924 × 3) / 100 = 27.72.
Hence, 25α = 25 × 27.72 = 693.

Final Answer: 693