Question

SHOW THAT MIDDLE TERM IN THE EXPANSION OF (1+X)^2n IS 1*3*5*7. . . .(2n-1)/2n!

Answer 1

Consider the expansion of $(1+x)^{2n}$. There are $2n+1$ terms in this expansion. The $r+1$ th term is

$T_{r+1}=\frac{(2n)!}{(2n-r)!r!} x^r$

The middle term is the $n+1$ the term. Hence $r=n$.

$T_{n+1}=\frac{(2n)!}{(2n-n)!n!} x^n\\ T_{n+1}=\frac{1*2*3*4*5*...(2n-1)*2n}{n!n!} x^n\\ T_{n+1}=\frac{1*3*5*...(2n-1)*2*4*...*2n}{n!n!} x^n\\ T_{n+1}=\frac{1*3*5*...(2n-1)*2^n(n!)}{n!n!} x^n\\ T_{n+1}=\frac{2^n*1*3*5*...(2n-1)}{n!} x^n$