Equal Middle Term Coefficients in Binomial Expansions

To find the value of $\alpha$, we need to determine the coefficients of the middle terms in both binomial expansions and equate them.

Step 1: Find the coefficient of the middle term in $(1 + \alpha x)^{26}$

Since $n = 26$ is an even integer, there is a single middle term given by the $\left(\frac{26}{2} + 1\right)\text{th} = 14\text{th}$ term. The general term in the expansion of $(1 + \alpha x)^{26}$ is: $T_{r+1} = \binom{26}{r} (\alpha x)^r$

For the $14\text{th}$ term, $r = 13$: $T_{14} = \binom{26}{13} (\alpha x)^{13}$

Thus, the coefficient of the middle term is: $C_1 = \binom{26}{13} \alpha^{13}$

Step 2: Find the coefficient of the middle term in $(1 - \alpha x)^{28}$

Since $n = 28$ is an even integer, the middle term is the $\left(\frac{28}{2} + 1\right)\text{th} = 15\text{th}$ term. The general term in the expansion of $(1 - \alpha x)^{28}$ is: $T_{r+1} = \binom{28}{r} (-\alpha x)^r$

For the $15\text{th}$ term, $r = 14$: $T_{15} = \binom{28}{14} (-\alpha x)^{14} = \binom{28}{14} \alpha^{14} x^{14}$

Thus, the coefficient of the middle term is: $C_2 = \binom{28}{14} \alpha^{14}$

Step 3: Equate the two coefficients and solve for $\alpha$

Given that $C_1 = C_2$: $\binom{26}{13} \alpha^{13} = \binom{28}{14} \alpha^{14}$

Since $\alpha \neq 0$, we can divide both sides by $\alpha^{13}$: $\alpha = \frac{\binom{26}{13}}{\binom{28}{14}}$

Expanding the combination terms: $\alpha = \frac{\frac{26!}{13! \, 13!}}{\frac{28!}{14! \, 14!}} = \frac{26!}{28!} \times \left(\frac{14!}{13!}\right)^2$

Simplifying the factorials: $\frac{26!}{28!} = \frac{1}{28 \times 27}$ $\frac{14!}{13!} = 14$

Substituting these back into the expression for $\alpha$: $\alpha = \frac{1}{28 \times 27} \times (14)^2 = \frac{196}{28 \times 27}$

Dividing the numerator and denominator by $28$: $\alpha = \frac{7}{27}$

Correct Answer: Option D ($\frac{7}{27}$)