Question

The ratio of fifth term from the beginning to the fifth term from the end in the expansion of ${(\sqrt[4]{42}+\frac{1}{\sqrt[4]{43}})}^n$ is $\sqrt{6}:1$. If $n=\frac{20}{\lambda }$, find the value of $\lambda$.

Answer 1

$T_{4+1}$ from begining is

${}^n{C}_4$ .$\frac{2^{\frac{n-4}{4}}}{3^{\frac{4}{4}}}$

And ${T"}_{4+1}$ from end

${}^n{C}_{n-4}$.$\frac{2^{\frac{4}{4}}}{3^{\frac{n-4}{4}}}$

$\Rightarrow$$\frac{T_5}{{T"}_5}$=$\sqrt{6}$

$\Rightarrow$$\frac{2^{\frac{n-4}{4}}.3^{\frac{n-4}{4}}}{2.3}$=$6^{\frac{1}{2}}$

$\Rightarrow$$\frac{6^{\frac{n-4}{4}}}{6}$=$6^{\frac{1}{2}}$

$\Rightarrow$$6^{\frac{n-4}{4}}$=$6^{\frac{3}{2}}$

hence

$\Rightarrow$ $\frac{n-4}{4}$=$\frac{3}{2}$

$\Rightarrow$ n-4=6 $\Rightarrow$n=10

$\Rightarrow$$\lambda$=$\frac{n}{2}$ so $\lambda$=2