Question

Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $(\sqrt[4]{42}+\frac{1}{\sqrt[4]{43}})$ is $\sqrt{6}:1$

Answer 1

We have $(\sqrt[4]{42}+\frac{1}{\sqrt[4]{43}})$

5th term from the beginning

$=n_{C_4}(\sqrt[4]{42}{)}^{n-4}{\left(\frac{1}{\sqrt[4]{43}}\right)}^4$

5th term from the end = (n+1-5+1)th term from beginning

$=(n-3)th$ term from the beginning

$=n_{C_{n-4}}(\sqrt[4]{42}{)}^4{\left(\frac{1}{\sqrt[4]{43}}\right)}^{n-4}$

Now $\frac{n_{C_4}(\sqrt[4]{42}{)}^{n-4}{\left(\frac{1}{\sqrt[4]{43}}\right)}^4}{n_{C_{n-4}}(\sqrt[4]{42}{)}^4{\left(\frac{1}{\sqrt[4]{43}}\right)}^{n-4}}=\frac{\sqrt{6}}{1}$

$\Rightarrow (2{)}^{\frac{n-8}{4}}.(3{)}^{\frac{n-8}{4}}=2^{\frac{1}{2}}\times 3^{\frac{1}{2}}$

$\Rightarrow \frac{n-8}{2}=\frac{1}{2}\Rightarrow n-8=2\Rightarrow n=10$