Question

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

Answer 1

The expression for the general term is
$T_{r+1}={}^n{C}_r{2}^{\frac{n-r}{4}}{.3}^{-\frac{r}{4}}$
Now 5th term from the beginning
$T_5$
$={}^n{C}_4{2}^{\frac{n-4}{4}}{.3}^{-\frac{4}{4}}$
$={}^n{C}_4{2}^{\frac{n-4}{4}}.\frac{1}{3}$ ...(i)
And the 5 th term from end is
$T_{n-3}$
$=T_{n-4+1}$
$={}^n{C}_{n-4}{2}^{\frac{n-(n-4)}{4}}{.3}^{-\frac{n-4}{4}}$
$={}^n{C}_4{2.3}^{-\frac{n-4}{4}}$ ...(ii)
Now
$\frac{i}{ii}=\frac{\sqrt{6}}{1}$
Or
$\frac{{}^n{C}_4{2}^{\frac{n-4}{4}}.\frac{1}{3}}{{}^n{C}_4{2.3}^{-\frac{n-4}{4}}}=\frac{\sqrt{6}}{1}$
$\frac{2^{\frac{n-4}{4}-1}}{3^{1-\frac{(n-4)}{4}}}=\sqrt{6}$
$(2\times 3{)}^{\frac{n-4}{4}-1}=\sqrt{6}$
$6^{\frac{n-8}{4}}=6^{\frac{1}{2}}$
Hence
$\frac{n-8}{4}=\frac{1}{2}$
$2n-16=4$
$2n=20$
$n=10$.