Question

In the expansion of \( (1 + \alpha x)^{26} \) and \( (1 - \alpha x)^{28} \), the coefficient of the middle term is the same, then the value of \( \alpha \) is:

• A. \( \dfrac{7}{22} \)
• B. \( \dfrac{7}{27} \)
• C. \( \dfrac{5}{27} \)
• D. \( \dfrac{5}{22} \)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a binomial expansion \( (a+b)^n \), if \( n \) is even, there is one middle term, which is the \( (\frac{n}{2} + 1)^{th} \) term. Its coefficient is \( \binom{n}{n/2} \). We equate the middle term coefficients for both given expansions.
Step 2: Key Formula or Approach:
1. Middle term of \( (1 + \alpha x)^{26} \) is \( T_{13+1} \). Coefficient: \( \binom{26}{13} \alpha^{13} \).
2. Middle term of \( (1 - \alpha x)^{28} \) is \( T_{14+1} \). Coefficient: \( \binom{28}{14} (-\alpha)^{14} = \binom{28}{14} \alpha^{14} \).
Step 3: Detailed Explanation:
Equating the two coefficients: \[ \binom{26}{13} \alpha^{13} = \binom{28}{14} \alpha^{14} \] \[ \alpha = \frac{\binom{26}{13}}{\binom{28}{14}} = \frac{26!}{(13! \cdot 13!)} \times \frac{(14! \cdot 14!)}{28!} \] \[ \alpha = \frac{26!}{28!} \times \frac{14 \times 14}{1} = \frac{1}{28 \times 27} \times 196 \] \[ \alpha = \frac{196}{756} \] Dividing both by 28: \[ \alpha = \frac{196 \div 28}{756 \div 28} = \frac{7}{27} \]
Step 4: Final Answer:
The value of \( \alpha \) is \( \frac{7}{27} \).