Find $x$ if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $(4\sqrt 2+\large\frac{1}{4\sqrt 3})^n$ is $\sqrt 6 :1$

Question

$\begin{array}{1 1}(A)\;8\\(B)\;9\\(C)\;10\\(D)\;11\end{array} $

Answer 1

Toolbox:

• $T_{r+1}$=General term =$nC_ra^{n-r}b^r$
• $\therefore (a+b)^n=\sum\limits_{r=0}^n nC_ra^{n-r}b^r$

$T_5$ in $\big[2^{\Large\frac{1}{4}}+\large\frac{1}{3^{\Large\frac{1}{4}}}\big]^n$$=nC_4(2^{1/4})^{n-4}.\big(\large\frac{1}{3^{1/4}}\big)^4$

$\Rightarrow nC_42^{\Large\frac{n-4}{4}}.\large\frac{1}{3}$-------(1)

Total number of terms =n+1

Fifth term from the end =$[(n+1)-5+1)]^{th}$ term from the beginning.

$\Rightarrow (n-3)^{th}$ term from the beginning.

$\Rightarrow nC_{n-4}(2^{1/4})^{n-(n-4)}\big(\large\frac{1}{3^{1/4}}\big)^{n-4}$

$\Rightarrow nC_4.2(\large\frac{1}{3})^{\Large\frac{n-4}{4}}$------(2)

Dividing (1) by (2)

$\large\frac{nC_42^{\Large\frac{n-4}{4}}.\large\frac{1}{3}}{nC_4.2(\large\frac{1}{3})^{\Large\frac{n-4}{4}}}=\frac{\sqrt 6}{1}$

Or $\large\frac{2^{\Large\frac{n}{4}-2}}{(1/3)^{\Large\frac{n}{4}-2}}=\frac{\sqrt 6}{1}$

Or $2^{\Large\frac{n}{4}-2}.3^{\Large\frac{n}{4}-2}=6^{\Large\frac{1}{2}}$

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

$\therefore \large\frac{n}{4}$$-2=\large\frac{1}{2}$

$\therefore \large\frac{n}{4}=\frac{5}{2}$

$n=\large\frac{5}{2}$$\times 4$

$n=10$

Hence (C) is the correct answer.