Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of (√[4]2+ 1 √[4]3)^n, in the increasing…

Given,

The ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of ${(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}})}^{n}$(42​+43​1​)n in the increasing powers of $\frac{1}{\sqrt[4]{3}}$43​1​ be $\sqrt[4]{6}:1$46​:1

Now according to given data we have,

$$\frac{{T}_{5}}{{T}_{n-3}}=\frac{C4n{({2}^{1/4})}^{n-4}{({3}^{-1/4})}^{4}}{Cn-4n{({2}^{1/4})}^{4}{({3}^{-1/4})}^{n-4}}=\frac{\sqrt[4]{6}}{1}$$Tn−3​T5​​=Cn−4n(21/4)4(3−1/4)n−4C4n(21/4)n−4(3−1/4)4​=146​​

$$\Rightarrow {2}^{\frac{n-8}{4}}{3}^{\frac{n-8}{4}}={6}^{1/4}$$⇒24n−8​34n−8​=61/4

$$\Rightarrow {6}^{n-8}=6$$⇒6n−8=6

$$\Rightarrow n-8=1\Rightarrow n=9$$⇒n−8=1⇒n=9

So, ${T}_{6}=C59{({2}^{1/4})}^{4}{({3}^{-1/4})}^{5}=\frac{84}{\sqrt[4]{3}}$T6​=C59(21/4)4(3−1/4)5=43​84​

Now comparing with ${T}_{6}=\frac{\alpha }{\sqrt[4]{3}}$T6​=43​α​,

We get $\alpha =84$α=84