Question

Show that the middle term of ${(x+\frac{1}{x})}^{4n}$ is equal to the coefficient of $x^n$ in the expansion of $(1-4x{)}^{-(n+\frac{1}{2})}$

Answer 1

Middle term of the expansion ${(x+\frac{1}{x})}^{4n}=\frac{4n!}{2n!}2n!$

$=\frac{2^{2n}2n!\cdot 1\cdot 3\cdot 5\dots (4n-1)}{2n!2n!}=2^{2n}\frac{1\cdot 3\cdot 5\dots (4n-1)}{2^{n}n!\cdot 1\cdot 3\cdot 5\dots (2n-1)}$

$=2^n\frac{(2n+1)(2n+3)\dots (4n-1)}{n!}$

$=\frac{4^n(n+\frac{1}{2})(n+\frac{3}{2})(n+\frac{5}{2})\dots \text{to n factors}}{n!}$

Coefficient of $x^n$ in the expansion $(1-4x{)}^{-(n+\frac{1}{2})}$ is

$\frac{(-n-\frac{1}{2})(-n-\frac{1}{2}-1)(-n-\frac{1}{2}-2)\dots \text{to n factors}}{n!}(-4{)}^n$

$=(-1)2n{4}^n\frac{(n+\frac{1}{2})(n+\frac{3}{2})(n+\frac{5}{2})\dots \text{to n factors}}{n!}$

$=4^n\frac{(n+\frac{1}{2})(n+\frac{3}{2})(n+\frac{5}{2})\dots \text{to n factors}}{n!}$

Hence Proved