JEE Main 2023 — Mathematics Algebra

Given expansion is${(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}})}^{n}$(42​+43​1​)n

We know that general term of an expansion ${(x+a)}^{n}$(x+a)n is ${T}_{r+1}=Crn{x}^{n-r}{a}^{r}$Tr+1​=Crnxn−rar(${5}^{\mathrm{th}}$5th term from beginning)

$${T}_{4+1}=C4n(\sqrt[4]{2}{)}^{n-4}\cdot {(\frac{1}{\sqrt[4]{3}})}^{4}$$T4+1​=C4n(42​)n−4⋅(43​1​)4

(${5}^{\mathrm{th}}$5th term from end)

We know that $r$rth term from the end in a binomial expansion is ${T}_{n-r+2}$Tn−r+2​ from the beginning.${T}_{5}^{'}={T}_{n-4+1}=C4n{(\frac{1}{\sqrt[4]{3}})}^{n-4}(\sqrt[4]{2}{)}^{4}$T5′​=Tn−4+1​=C4n(43​1​)n−4(42​)4

It is also given that the ratio of the fifth term from the beginning to the fifth term from the end is $\sqrt{6}:1$6​:1.

Now, $\frac{{T}_{5}}{{T}_{5}^{'}}=\frac{\sqrt{6}}{1}$T5′​T5​​=16​​

$$\Rightarrow {2}^{\frac{n-8}{4}}\cdot {3}^{\frac{n-8}{4}}=\sqrt{6}$$⇒24n−8​⋅34n−8​=6​

$$\therefore (\sqrt{6}{)}^{\frac{n-8}{2}}=(\sqrt{6})$$∴(6​)2n−8​=(6​)

$$\therefore n=10$$∴n=10

$$\therefore {T}_{3}=C210{(\sqrt[4]{2})}^{8}{(\frac{1}{\sqrt[4]{3}})}^{2}=\frac{45\times 4}{\sqrt{3}}$$∴T3​=C210(42​)8(43​1​)2=3​45×4​

$$=60\sqrt{3}$$=603​

Hence, this is the correct option.