Show that the middle term in the expansion of $(1+x)^{2 n}$ is
$\frac{1.3 .5 \ldots(2 n-1)}{n !} 2 n\, x^{n},$ where $n$ is a positive integer.

Let the sixth term in the binomial expansion of $\left(\sqrt{2^{\log _2}\left(10-3^x\right)}+\sqrt[5]{2^{(x-2) \log _2 3}}\right)^m$, in the increasing powers of $2^{(x-2) \log _2 3}$, be $21$ . If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an $A.P.$, then the sum of the squares of all possible values of $x$ is $………$.

Find the cocfficient of $a^{5} b^{7}$ in $(a-2 b)^{12}$

Let $\mathrm{m}$ and $\mathrm{n}$ be the coefficients of seventh and thirteenth terms respectively in the expansion of $\left(\frac{1}{3} \mathrm{x}^{\frac{1}{3}}+\frac{1}{2 \mathrm{x}^{\frac{2}{3}}}\right)^{18}$. Then $\left(\frac{\mathrm{n}}{\mathrm{m}}\right)^{\frac{1}{3}}$ is :

Coefficient of $x^{11}$ in the expansion of $\left(1+x^2\right)^4\left(1+x^3\right)^7\left(1+x^4\right)^{12}$ is