Given, expansion

$$\begin{aligned} & =\left(1+x+x^2+x^3\right)^{11}=\left[(1+x)+x^2(1+x)\right]^{11} \\ & =\left[(1+x)\left(1+x^2\right)\right]^{11}=(1+x)^{11} \cdot\left(1+x^2\right)^{11} \end{aligned}$$

Now, above expansion becomes

$$\begin{aligned} & =\left({ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^2+{ }^{11} C_3 x^3+{ }^{11} C_4 x^4+\ldots\right)\left({ }^{11} C_0+{ }^{11} C_1 x^2+{ }^{11} C_2 x^4+\ldots\right) \\ & =\left(1+11 x+55 x^2+165 x^3+330 x^4+\ldots\right)\left(1+11 x^2+55 x^4+\ldots\right) \end{aligned}$$

$\therefore$ Coefficient of $x^4=55+605+330=990$

Given expansion is $\left(\frac{p}{2}+2\right)^8$.

Here, $n=8\quad$ [even]

Since, this Binomial expansion has only one middle term i.e., $\left(\frac{8}{2}+1\right)$ th $=5$ th term

$$\begin{array}{l} & T_5 & =T_{4+1}={ }^8 C_4\left(\frac{p}{2}\right)^{8-4} \cdot 2^4 \\ \Rightarrow \quad & 1120 & ={ }^8 C_4 p^4 \cdot 2^{-4} 2^4 \\ \Rightarrow \quad & 1120 & =\frac{8 \times 7 \times 6 \times 5 \times 4!}{4!\times 4 \times 3 \times 2 \times 1} p^4 \end{array}$$

$$\begin{aligned} & \Rightarrow \quad 1120=7 \times 2 \times 5 \times p^4 \\ & \Rightarrow \quad p^4=\frac{1120}{70}=16 \Rightarrow p^4=2^4 \\ & \Rightarrow \quad p^2=4 \Rightarrow p= \pm 2 \end{aligned}$$

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.