If the coefficients of the middle terms in the binomial expansions of (1+ x)^26 and (1- x)…

Question

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

Accepted Answer

For a binomial expansion $(a+b{)}^{n}$:

if $n$ is even, there is only one middle term,

and the middle term is the $(\frac{n}{2}+1)$th term.

We are given:

$(1+\alpha x{)}^{26}\, \, (1-\alpha x{)}^{28}$

and the coefficients of their middle terms are equal.

1. Middle term of $(1+\alpha x{)}^{26}$
Since $26$ is even, the middle term is the $(\frac{26}{2}+1)=14$th term.

The general term in $(1+\alpha x{)}^{26}$ is

${T}_{r+1}=(\frac{26}{r})(1{)}^{26-r}(\alpha x{)}^{r}$

So the $14$th term corresponds to $r=13$:

${T}_{14}=(\frac{26}{13})(\alpha x{)}^{13}$

Hence, the coefficient of the middle term is

$(\frac{26}{13}){\alpha }^{13}$

2. Middle term of $(1-\alpha x{)}^{28}$
Since $28$ is even, the middle term is the $(\frac{28}{2}+1)=15$th term.

The general term is

${T}_{r+1}=(\frac{28}{r})(1{)}^{28-r}(-\alpha x{)}^{r}$

So the $15$th term corresponds to $r=14$:

${T}_{15}=(\frac{28}{14})(-\alpha x{)}^{14}$

Since $14$ is even,

$(-\alpha x{)}^{14}={\alpha }^{14}{x}^{14}$

Thus, the coefficient of the middle term is

$(\frac{28}{14}){\alpha }^{14}$

3. Equate the coefficients
Given that these coefficients are equal:

$(\frac{26}{13}){\alpha }^{13}=(\frac{28}{14}){\alpha }^{14}$

Since $\alpha 
e 0$, divide both sides by ${\alpha }^{13}$:

$(\frac{26}{13})=(\frac{28}{14})\alpha $

So,

$\alpha =\frac{(\frac{26}{13})}{(\frac{28}{14})}$

4. Simplify
$\alpha =\frac{\frac{26!}{13!13!}}{\frac{28!}{14!14!}}$

$\alpha =\frac{26!}{13!13!}\cdot \frac{14!14!}{28!}$

Now use

$14!=14\cdot 13!$

So,

$14!14!=(14\cdot 13!)(14\cdot 13!)={14}^{2}(13!{)}^{2}$

Thus,

$\alpha =\frac{26!\cdot {14}^{2}(13!{)}^{2}}{(13!{)}^{2}\cdot 28!}=\frac{26!\cdot {14}^{2}}{28!}$

Also,

$28!=28\cdot 27\cdot 26!$

Therefore,

$\alpha =\frac{{14}^{2}}{28\cdot 27}=\frac{14}{2\cdot 27}=\frac{7}{27}$

$\alpha =\frac{7}{27}$

So, the correct option is D.

Solutions · 1