Question

If the ratio of the 5th term from the beginning to the 5th term from the end in the expansion of ${(4\sqrt{2}+\frac{1}{4\sqrt{3}})}^n$
is $\sqrt{6}:1$ then find the value of n.

Answer 1

The given expression may be written as $(2^{\frac{1}{4}}+3^{-\frac{1}{4}}{)}^n$
The general term in this expansion is given by
$T_{r+1}{=}^n{C}_r\times (2^{\frac{1}{4}}{)}^{(n-r)}\times (3^{-\frac{1}{4}}{)}^r\Rightarrow T_{r+1}{=}^n{C}_r\times 2^{\frac{(n-r)}{4}}\times 3^{\left(\frac{-r}{4}\right)}$
Now, pth term from the end
= (n- p + 2)th term from the beginning.
$\therefore$ 5th term from the end
= (n - 5 + 2)th term from the beginning
= (n - 3)th term from the beginning.
Now, $T_5=T_{(4+1)}{=}^N{C}_4\times 2^{\frac{(n-4)}{4}}\times 3^{\left(\frac{-4}{4}\right)}{=}^n{C}_4\times 2^{\frac{(a-4)}{4}}\times 3^{-1}\text{And,}{T}_{n-3}=T_{(n-4)+1}{=}^n{C}_{n-4}\times 2^1\times 3^{\frac{(n-4)}{4}}{=}^n{C}_4\times 2\times 3^{\frac{-(n-4)}{4}}\therefore \frac{T_5}{T_{n-3}}=\frac{\sqrt{6}}{1}\Rightarrow \frac{{}^n{C}_4\times 2^{\frac{(n-4)}{4}\times 3^{-1}}}{{}^n{C}_4\times 2\times 3^{\frac{-(n-4)}{4}}}=\frac{\sqrt{6}}{1}\Rightarrow 2^{|\frac{(n-4)}{4}-1|}\times 3^{\{\frac{(n-4)}{4}-1\}}=\sqrt{6}\Rightarrow 2^{\frac{(n-8)}{4}}\times 3^{\frac{(n-8)}{4}}=6^{\frac{1}{2}}\Rightarrow (2\times 3{)}^{\frac{(n-8)}{4}}=6^{\frac{1}{2}}\Rightarrow \frac{n-8}{4}=\frac{1}{2}\Rightarrow 2n-16=4\Rightarrow n=10$