Question

If the coefficients of the middle terms in the binomial expansions of $(1 + \alpha x)^{26}$ and $(1 - \alpha x)^{28}$, $\alpha \neq 0$, are equal, then the value of $\alpha$ is:

• A. 1
• B. $$\frac{14}{13}$$
• C. $$\frac{27}{7}$$
• D. $$\frac{7}{27}$$

Hint:

Identify that for an even power $n$, the middle term is the $(n/2 + 1)^{th}$ term. Find the general term, extract the coefficients, and solve for alpha by equating them.

Answer 1

The Correct Option is D

In a binomial expansion $(a + b)^n$, if $n$ is even, there is one middle term, which is the $(\frac{n}{2} + 1)^{th}$ term.

1. Middle term of $(1 + \alpha x)^{26}$:
Here $n=26$. The middle term is $T_{\frac{26}{2} + 1} = T_{13+1} = T_{14}$.
General term formula: $T_{r+1} = ^nC_r a^{n-r} b^r$.
$T_{14} = ^{26}C_{13} (1)^{13} (\alpha x)^{13} = ^{26}C_{13} \alpha^{13} x^{13}$.
Coefficient of this middle term is $C_1 = ^{26}C_{13} \alpha^{13}$.

2. Middle term of $(1 - \alpha x)^{28}$:
Here $n=28$. The middle term is $T_{\frac{28}{2} + 1} = T_{14+1} = T_{15}$.
$T_{15} = ^{28}C_{14} (1)^{14} (-\alpha x)^{14} = ^{28}C_{14} (-\alpha)^{14} x^{14} = ^{28}C_{14} \alpha^{14} x^{14}$.
Coefficient of this middle term is $C_2 = ^{28}C_{14} \alpha^{14}$.

3. Equating the coefficients:
Given $C_1 = C_2$, we have:
$$^{26}C_{13} \alpha^{13} = ^{28}C_{14} \alpha^{14}$$
Since $\alpha \neq 0$, we can divide both sides by $\alpha^{13}$:
$$\alpha = \frac{^{26}C_{13}}{^{28}C_{14}}$$

4. Simplifying the ratio of combinations:
Using $^nC_r = \frac{n!}{r!(n-r)!}$, we get:
$$\alpha = \frac{26!}{13!13!} \times \frac{14!14!}{28!}$$
$$\alpha = \frac{26!}{28!} \times \frac{14!}{13!} \times \frac{14!}{13!}$$
$$\alpha = \frac{1}{28 \times 27} \times 14 \times 14$$
$$\alpha = \frac{196}{756} = \frac{14}{2 \times 27} = \frac{7}{27}$$.