Given, expansion is $\left(x-\frac{1}{x}\right)^{2 n}$. This Binomial expansion has even power. So, this has one middle term.

$$\begin{aligned} \text{i.e.,}\quad \left(\frac{2 n}{2}+1\right) \text { th term }=(n+1) \text { th term } \\ T_{n+1} & ={ }^{2 n} C_n(x)^{2 n-n}\left(-\frac{1}{x}\right)^n={ }^{2 n} C_n x^n(-1)^n x^{-n} \\ & ={ }^{2 n} C_n(-1)^n=(-1)^n \frac{(2 n)!}{n!n!}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \ldots(2 n-1)(2 n)}{n!n!}(-1)^n \\ & =\frac{1 \cdot 3 \cdot 5 \ldots(2 n-1) \cdot 2 \cdot 4 \cdot 6 \ldots(2 n)}{12 \cdot 3 \cdot \ldots n(n!)}(-1)^n \\ & =\frac{1 \cdot 3 \cdot 5 \ldots(2 n-1) \cdot 2^n(1 \cdot 2 \cdot 3 \ldots n)(-1)^n}{(1 \cdot 2 \cdot 3 \ldots n)(n!)} \\ & =\frac{[1 \cdot 3 \cdot 5 \ldots(2 n-1)]}{n!}(-2)^n \quad \text{Hence proved.} \end{aligned}$$

Here, the Binomial expansion is $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n$.

Now, 7 thterm from beginning $T_7=T_{6+1}={ }^n C_6(\sqrt[3]{2})^{n-6}\left(\frac{1}{\sqrt[3]{3}}\right)^6\quad \text{... (i)}$

and 7 th term from end i.e., $T_7$ from the beginning of $\left(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{2}\right)^n$

i.e., $$T_7={ }^n C_6\left(\frac{1}{\sqrt[3]{3}}\right)^{n-6}(\sqrt[3]{2})^6\quad \text{.... (ii)}$$

Given that, $\frac{{ }^n C_6(\sqrt[3]{2})^{n-6}\left(\frac{1}{\sqrt[3]{3}}\right)^6}{{ }^n C_6\left(\frac{1}{\sqrt[3]{3}}\right)^{n-6}(\sqrt[3]{2})^6}=\frac{1}{6} \Rightarrow \frac{2^{\frac{n-6}{3}} \cdot 3^{-6 / 3}}{3^{-\left(\frac{n-6}{3}\right)} \cdot 2^{6 / 3}}=\frac{1}{6}$

$\Rightarrow\left(2^{\frac{n-6}{3}} \cdot 2^{\frac{-6}{3}}\right)\left(3^{\frac{-6}{3}} \cdot 3^{\frac{(n-6)}{3}}\right)=6^{-1}$

$\begin{array}{rlrl}\Rightarrow & \left(2^{\frac{n-6}{3}-\frac{6}{3}}\right) \cdot\left(3^{\frac{n-6}{3}-\frac{6}{3}}\right) & =6^{-1} \Rightarrow(2 \cdot 3)^{\frac{n}{3}-4}=6^{-1} \\ \Rightarrow & \frac{n}{3}-4 & =-1 \Rightarrow \frac{n}{3}=3 \\ \therefore \quad n & =9\end{array}$

(i) Given expansion is $(x+a)^n$.

$$\therefore(x+a)^n={ }^n C_0 x^n a^0+{ }^n C_1 x^{n-1} a^1+{ }^n C_2 x^{n-2} a^2+{ }^n C_3 x^{n-3} a^3+\ldots+{ }^n C_n a^n$$

Now, sum of odd terms

i.e., $$O={ }^n C_0 x^n+{ }^n C_2 x^{n-2} a^2+\ldots$$

and sum of even terms

i.e., $$E={ }^n C_1 x^{n-1} a+{ }^n C_3 x^{n-3} a^3+\ldots$$

$$\begin{aligned} \because \quad & (x+a)^n =O+E \quad \text{.... (i)}\\ \text { Similarly, } \quad & (x-a)^n =O-E & \ldots \text { (ii) } \\ \therefore \quad & (O+E)(O-E) & =(x+a)^n(x-a)^n & \text { [on multiplying Eqs. (i) and (ii)] } \\ \Rightarrow \quad & O^2-E^2 & =\left(x^2-a^2\right)^n & \end{aligned}$$

(ii) $4 O E=(O+E)^2-(O-E)^2=\left[(x+a)^n\right]^2-\left[(x-a)^n\right]^2\quad \text{[from Eqs. (i) and (ii)] }$

$$=(x+a)^{2 n}-(x-a)^{2 n}$$ Hence proved.

Given expansion is $${\left( {{x^2} + {1 \over x}} \right)^{2n}}$$.

Let $x^p$ occur in the expansion of $${\left( {{x^2} + {1 \over x}} \right)^{2n}}$$.

$${T_{r + 1}} = {}^{2n}{C_r}{({x^2})^{2n - r}}{\left( {{1 \over x}} \right)^r}$$

$$ = {}^{2n}{C_r}{x^{4n - 2r}}{x^{ - r}} = {}^{2n}{C_r}{x^{4n - 3r}}$$

Let $$4n - 3r = p$$

$$ \Rightarrow 3r = 4n - p \Rightarrow r = {{4n - p} \over 3}$$

$\therefore$ Coefficient of $${x^p} = {}^{2n}{C_r} = {{(2n)!} \over {r!(2n - r)!}} = {{(2n)!} \over {\left( {{{4n - p} \over 3}} \right)!\left( {2n - {{4n - p} \over 3}} \right)!}}$$

$$ = {{(2n)!} \over {\left( {{{4n - p} \over 3}} \right)!\left( {{{6n - 4n + p} \over 3}} \right)!}} = {{(2n)!} \over {\left( {{{4n - p} \over 3}} \right)!\left( {{{2n + p} \over 3}} \right)!}}$$

Given expansion is $\left(1+x+2 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$

Now, consider $\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$

$$\begin{aligned} T_{r+1} & ={ }^9 C_r\left(\frac{3}{2} x^2\right)^{9-r}\left(-\frac{1}{3 x}\right)^r \\ & ={ }^9 C_r\left(\frac{3}{2}\right)^{9-r} x^{18-2 r}\left(-\frac{1}{3}\right)^r x^{-r}={ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{18-3 r} \end{aligned}$$

Hence, the general term in the expansion of $\left(1+x+2 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$

$$={ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{18-3 r}+{ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{19-3 r}+2 \cdot{ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{21-3 r}$$

For term independent of $x$, putting $18-3 r=0,19-3 r=0$ and $21-3 r=0$, we get

$$r=6, r=19 / 3, r=7$$

Since, the possible value of $r$ are 6 and 7.

Hence, second term is not independent of $x$.

$$\begin{aligned} & =\frac{9 \times 8 \times 7 \times 6!}{6!\times 3 \times 2} \cdot \frac{3^3}{2^3} \cdot \frac{1}{3^6}-2 \cdot \frac{9 \times 8 \times 7!}{7!\times 2 \times 1} \cdot \frac{3^2}{2^2} \cdot \frac{1}{3^7} \\ & =\frac{84}{8} \cdot \frac{1}{3^3}-\frac{36}{4} \cdot \frac{2}{3^5}=\frac{7}{18}-\frac{2}{27}=\frac{21-4}{54}=\frac{17}{54} \end{aligned}$$