Question

Find the coefficient of the middle term in the expansion of $(1+x{)}^n$ when n is even and the coefficient of middle terms if n is odd. Your answer must not be in factorial notations.

Answer 1

If n is even, then middle term is [(n + 2) + 1]th. Hence
$T_{n/2+1}={}^n{C}_{n/2}{x}^{n/2}$
$\therefore$ coefficient is ${}^n{C}_{n/2}=\frac{n!}{\left(n/2\right)!\cdot \left(n/2\right)!}$ (n is even)
$=\frac{n(n-1)(n-2).....3\cdot 2\cdot 1}{\left(n/2\right)!\cdot \left(n/2\right)!}$
= [n (n - 2).....4.2] [(n - 1) (n - 3) .... 3.1] $÷$ (n / 2)! . (n / 2)!
Each bracket contains n / 2 numbers and the first contains all even numbers and 2nd all odd. You can take 2 common from each of the n / 2 even numbers.
$=2^{n/2}[\frac{n}{2}(\frac{n}{2}-1).....2.1]$
$\left[\right(n-1\left)\right(n-3)....5\cdot 3\cdot 1]÷\left(n/2\right)!\cdot \left(n/2\right)!$
$=2^{n/2}\cdot \left(n/2\right)!\left[\right(n-1\left)\right(n-3)...5\cdot 3\cdot 1]÷\left(n/2\right)!\cdot \left(n/2\right)!$
$=2^{n/2}!\frac{\left[\right(n-1\left)\right(n-3).....5\cdot 3\cdot 1]}{[1\cdot 2\cdot 3.......n/2]}$
2nd part : If n is odd, then the two middle terms are
$\left(\frac{n+1}{2}\right)thand\left(\frac{n+3}{2}\right)th$
i.e. $T_{(n+1)/2}=T_{(n-1)/2+1}={}^n{C}_{(n-1)/2}{x}^{(n-1)/2}$
and $T_{(n+3)/2}=T_{(n+1)/2+1}={}^n{C}_{(n+1)/2}{x}^{(n+1)/2}$
$\therefore$ The coefficients are
$Butas{}^n{C}_r={}^n{C}_{n-r}$ therefore the above two coefficients are equal. If should be noted when n is odd then both
$\frac{n-1}{2}abnd\frac{n+1}{2}$ are integers.
Now proceeding as above the coefficients is
$2^{(n-1)/2}\frac{1\cdot 3\cdot 5.......n}{1\cdot 2............\left[\right(n+1\left)/2\right]}$