Question

In the expansion of $(2^\frac{1}{4}+3^{-\frac{1}{4}})^n$, the ratio of $5^{th}$ term from start and $5^{th}$ term form end is $\sqrt 6 : 1$ , then find $3^{rd}$ term

• A. 30$\sqrt 3$
• B. 60$\sqrt 3$
• C. 30
• D. 50$\sqrt 3$

Answer 1

Answer: (B)

60$\sqrt 3$

Answer 2

$\frac{^nC_4 (2^\frac{1}{4})^{n-4}(3^{\frac{-1}{4}})^4}{^nC_4 (2^\frac{-1}{4})^{n-4}(3^{\frac{1}{4}})^4}$ = $\sqrt 6$
$(\frac{2^\frac{1}{4}}{3^\frac{-1}{4}})^{(n-8)}$ = $\sqrt 6$
$(6)^\frac{n-8}{4}$ = $\sqrt 6$
$n-8=2$
$n=10$
$T_3 = {^{10}}C_2(2^{\frac{1}{4}})^8 (3^\frac{-1}{4})^2$
$= {^{10}}C_2\times(\sqrt2)^4\times\frac{1}{\sqrt3} = 60\sqrt3$

So, the correct answer is (B): $60\sqrt 3$