Question

If the ratio of the $5^{\text{th}}$ term from the beginning to the $5^{\text{th}}$ term from the end in the expansion of ${(\sqrt[4]{42}+\frac{1}{\sqrt[4]{43}})}^n$ is $\sqrt{6}:1,$ then the value of $n$ is

• A. $12$
• B. $10$
• C. $8$
• D. $14$

Answer 1

The correct option is B $10$

The given expression may be written as ${(2^{\frac{1}{4}}+3^{-\frac{1}{4}})}^n.$
The general term in this expansion is given by $T_{r+1}={}^n{C}_r\times {\left(2^{\frac{1}{4}}\right)}^{(n-r)}\times {\left(3^{-\frac{1}{4}}\right)}^r$
$\Rightarrow T_{r+1}={}^n{C}_r\times 2^{\frac{(n-r)}{4}}\times {\left(3^{-\frac{1}{4}}\right)}^r...\left(1\right)$
Now, $p^{\text{th}}$ term from the end
$=(n-p+2{)}^{\text{th}}$ term from the beginning.
$\therefore 5^{\text{th}}$ term from the end
$=(n-5+2{)}^{\text{th}}$ term from the beginning
$=(n-3{)}^{\text{th}}$ term from the beginning

Now, $T_5=T_{4+1}$
$={}^n{C}_4\times 2^{\frac{(n-4)}{4}}\times 3^{\left(\frac{-4}{4}\right)}$
$={}^n{C}_4\times 2^{\frac{(n-4)}{4}}\times 3^{-1}...\left(2\right)$
And, $T_{n-3}=T_{(n-4)+1}$
$={}^n{C}_{n-4}\times 2^1\times 3^{\frac{-(n-4)}{4}}$
$($putting $r=(n-4)$ in eqn $\left(1\right))$

$={}^n{C}_4\times 2\times 3^{\frac{-(n-4)}{4}}...\left(3\right)$
$[\because {}^n{C}_{n-r}={}^n{C}_r]$
$\therefore \frac{T_5}{T_{n-3}}=\frac{\sqrt{6}}{1}$
$\Rightarrow \frac{{}^n{C}_4\times 2^{\frac{n-4}{4}}\times 3^{-1}}{{}^n{C}_4\times 2\times 3^{\frac{-(n-4)}{4}}}=\frac{\sqrt{6}}{1}$
$\Rightarrow 2^{(\frac{(n-4)}{4}-1)}\times 3^{(\frac{n-4}{4}-1)}=\sqrt{6}$
$\Rightarrow (2\times 3{)}^{\frac{(n-8)}{4}}=6^{\frac{1}{2}}$
$\Rightarrow \frac{(n-8)}{4}=\frac{1}{2}$
$\Rightarrow n-8=2\Rightarrow n=10.$
Hence, $n=10$.