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

We know the formula for the middle term coefficient of $(1+x)^n$ for even $n$ is $^nC_{n/2}$. For $(1+kx)^n$, the coefficient is $^nC_{n/2} \cdot k^{n/2}$.

For the first expression $(1 + \alpha x)^{26}$, $n=26$ and $k=\alpha$. Coefficient is $C_{m1} = ^{26}C_{13} \alpha^{13}$.
For the second expression $(1 - \alpha x)^{28}$, $n=28$ and $k=-\alpha$. Coefficient is $C_{m2} = ^{28}C_{14} (-\alpha)^{14} = ^{28}C_{14} \alpha^{14}$.

Setting them equal:
$^{26}C_{13} \alpha^{13} = ^{28}C_{14} \alpha^{14}$
$\alpha = \frac{^{26}C_{13}}{^{28}C_{14}}$

Let's use the property $^nC_r = \frac{n}{r} \cdot ^{n-1}C_{r-1}$ to relate the two coefficients:
$^{28}C_{14} = \frac{28}{14} \cdot ^{27}C_{13} = 2 \cdot ^{27}C_{13}$
And $^{27}C_{13} = \frac{27}{14} \cdot ^{26}C_{13}$ (since $^nC_r = \frac{n}{n-r} \cdot ^{n-1}C_r$)
So, $^{28}C_{14} = 2 \cdot \frac{27}{14} \cdot ^{26}C_{13} = \frac{27}{7} \cdot ^{26}C_{13}$

Substitute this back into the expression for $\alpha$:
$\alpha = \frac{^{26}C_{13}}{\frac{27}{7} \cdot ^{26}C_{13}} = \frac{1}{27/7} = \frac{7}{27}$.