Question

Show that the middle term in the expansion of $(1+x{)}^{2n}$ is $\frac{1.3.5\cdots (2n-1)}{n!}.2^n.X^n,\text{where}nϵN.$

Answer 1

Clearly, the number of terms in th expansion of$(1+x{)}^{2n}$ is (2n + 1).
$\therefore \text{middle term}=(\frac{2n}{2}+1)\text{th term = (n + 1)th term}{T}_{n+1}\text{Now},T_{n+1}{=}^{2n}{C}_n.X^n=\frac{\left(2n\right)!}{(n!)\times (n!)}.x^n=\frac{1.2.3.4\cdots (2n-2)(2n-1)\left(2n\right)}{(n!)\times (n!)}.x^n=\frac{[1.3.5\cdots (2n-3\left)\right(2n-1\left)\right]\times [\mathrm{2.4.6}\cdots (2n-2\left)\right(2n\left)\right]}{(n!)(n!)}.x^n=\frac{[\mathrm{1.3.5}\cdots (2n-1\left)\right]\times 2^n[\mathrm{1.2.3}\cdots (n-1).n]}{(n!)\times (n!)}.x^n=\frac{[\mathrm{1.3.5}\cdots (2n-1\left)\right]\times 2^n\times x^n}{(n!)}$
Hence the middle term in the given expansion is
$\frac{\mathrm{1.3.5}\cdots (2n-1)\times 2^n\times x^n}{(n!)}$