Question

If coefficients of middle terms in the expansion \( (1 + \alpha x)^{26}\) & \( (1 - \alpha x)^{28} \) are equal then \( \alpha \) is:

• A. \( \frac{1}{4} \)
• B. \( \frac{8}{27} \)
• C. \( \frac{7}{27} \)
• D. \( \frac{9}{28} \)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In an expansion \( (a+b)^n \), if \( n \) is even, the middle term is the \( (\frac{n}{2} + 1)^{th} \) term. We equate the coefficients of these terms for both binomials.

Step 2: Key Formula or Approach:
1. Middle term of \( (1 + \alpha x)^{26} \) is \( T_{14} \), coefficient is \( \binom{26}{13} \alpha^{13} \). 2. Middle term of \( (1 - \alpha x)^{28} \) is \( T_{15} \), coefficient is \( \binom{28}{14} (-\alpha)^{14} = \binom{28}{14} \alpha^{14} \).

Step 3: Detailed Explanation:
1. Equate coefficients: \[ \binom{26}{13} \alpha^{13} = \binom{28}{14} \alpha^{14} \implies \alpha = \frac{\binom{26}{13}}{\binom{28}{14}} \] 2. Expand the combinations: \[ \alpha = \frac{26!}{13!13!} \times \frac{14!14!}{28!} = \frac{26!}{28!} \times \frac{14!14!}{13!13!} \] \[ \alpha = \frac{1}{28 \times 27} \times (14 \times 14) \] 3. Simplify: \[ \alpha = \frac{196}{756} = \frac{14 \times 14}{2 \times 14 \times 27} = \frac{14}{2 \times 27} = \frac{7}{27} \]

Step 4: Final Answer:
The value of \( \alpha \) is \( \frac{7}{27} \).