Question

if the ratio of the fifth term form the beginning to the fifth 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. $8$
• B. $9$
• C. $10$
• D. $11$

Answer 1

The correct option is C $10$

${(\sqrt[4]{42}+\frac{1}{\sqrt[4]{43}})}^n$

$5^{\text{th}}\text{term from beginning}$

${}^n{C}_4\text{(2)}(3{)}^{\frac{n-4}{4}}$

$5^{\text{th}}\text{term from end}(n-5+2)=(n-3{)}^{2n}$

${}^n{C}_{n-4}\left(2{)}^{\frac{n-4}{4}}\right(3)$

$\text{Now}\frac{{}^n{C}_4\left(2\right)(3{)}^{\frac{n-4}{4}}}{{}^n{C}_{n-4}\left(2{)}^{\frac{n-4}{4}}\right(3)}=\sqrt{6}$

$\left(2{)}^{(1-\frac{n}{4}+1)}\right(3{)}^{(\frac{n}{4}-2)}=\sqrt{6}$

$\left(2{)}^{2-n/4}\right(3{)}^{\frac{n}{4}-2}=3^{1/2}\times 2^{1/2}$

$\therefore n=10$