Question

Find n, if the ratio of the fifth term from the beginning to fifth term from the end in the expansion of ${(\sqrt[4]{42}+\frac{1}{\sqrt[4]{43}})}^n$ is $\sqrt{6}:1.$

Or

Prove that the coeffficient of the middle term in the expansion of $(1+x{)}^{2n}$ is equal to the sum of the coefficient of middle terms in the expansion of $(1+x{)}^{2n-1}.$

Answer 1

We have, ${(\sqrt[4]{42}+\frac{1}{\sqrt[4]{43}})}^n$

Now, 5th term from beginnning,

i.e. $T_5={}^n{C}_r(\sqrt[4]{42}{)}^{n-4}\times {\left(\frac{1}{\sqrt[4]{43}}\right)}^2$

= ${}^n{C}_4(2{)}^{\frac{n-4}{4}}\times \frac{1}{(3{)}^{4/4}}={}^n{C}_4\frac{(2{)}^{\frac{n-4}{4}}}{3}$

and 5th term from end, i.e.

$T_5^{\prime }$ from the beginning of ${(\frac{1}{\sqrt[4]{43}}+\sqrt[4]{42})}^n$

$\Rightarrow T_5^{\prime }={}^n{C}_4{\left(\frac{1}{\sqrt[4]{43}}\right)}^{n-4}(\sqrt[4]{42}{)}^4={}^n{C}_4\frac{1}{(3{)}^{\frac{n-4}{4}}}\times (2{)}^{4/4}$

= ${}^n{C}_4(3{)}^{\frac{4-n}{4}}\times 2$

Given that $\frac{T_5}{T_5^{\prime }}=\frac{\sqrt{6}}{1}$

$\Rightarrow \frac{{}^n{C}_4(2{)}^{\frac{n-4}{4}}}{3{\times }^n{C}_4(3{)}^{\frac{4-n}{4}}\times 2}=\frac{\sqrt{6}}{1}$

$\Rightarrow (2{)}^{\frac{n-4}{4}}.(3{)}^{\frac{n-4}{4}}=6\sqrt{6}$

$\Rightarrow (6{)}^{\frac{n-4}{4}}={6.6}^{\frac{1}{2}}$

$\Rightarrow (6{)}^{\frac{n-4}{4}}=(6{)}^{\frac{3}{2}}$

On comparing the powers, we get

$\frac{n-4}{4}=\frac{3}{2}\Rightarrow n-4=6$

$\therefore n=10$

Or

The middle term in the expansion of $(1+x{)}^{2n}$ is given by

$T_{n+1}={}^{2n}{C}_n{x}^n$

So, the coefficient of the middle term in the expansion of $(1+x{)}^{2n}$ is ${}^{2n}{C}_n.$

Now, the mioddle terms in the expansion of $(1+x{)}^{2n-1}$ is given by

$T_n$ and $T_{n+1}$ $[\because 2n-1isodd]$

Now, $T_n={}^{2n-1}{C}_{n-1}{x}^{n-1}$

and $T_{n+1}={}^{2n-1}{C}_n{x}^n$

So, the coefficient of the middle terms in the expansion of $(1+x{)}^{2n-1}$ are ${}^{2n-1}{C}_{n-1}$ and ${}^{2n-1}{C}_n.$

$\therefore$ Sum of these coefficient

= ${}^{2n-1}{C}_{n-1}{+}^{2n-1}{C}_n$

= ${}^{2n-1+1}{C}_n[\because {}^n{C}_{r-1}{+}^n{C}_r={}^{n+1}{C}_r]$

= ${}^{2n}{C}_{n^{\prime }}$

which is the coefficient of middle term in the expansion of $(1+x{)}^{2n}.$
Hence proved.