Question

Show that the middle term in the expansion of ${{(1+x)}^{2n}}$ is $\dfrac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{n!}\cdot {{2}^{n}}{{x}^{n}}$ , where $n\in \mathbb{N}$.

Answer 1

We will compare the given expansion of ${{(1+x)}^{2n}}$with ${{(a+b)}^{k}}$. We will find the index of the term that is the middle term. Then, we will use the formula of the general term of the expansion to find the middle term. Simplifying this term will lead us to the desired answer.

Complete step by step answer:
For the expansion of ${{(a+b)}^{k}}$, the middle term is the ${{\left( \dfrac{k}{2}+1 \right)}^{th}}$ term if $k$ is even. Comparing ${{(1+x)}^{2n}}$ with ${{(a+b)}^{k}}$, we have $a=1$, $b=x$, $k=2n$. The index of the middle term for ${{(1+x)}^{2n}}$ is $\left( \dfrac{2n}{2}+1 \right)=n+1$.
The formula for the general term of ${{(a+b)}^{k}}$ is as follows,
${{T}_{r+1}}={}^{k}{{C}_{r}}\cdot {{a}^{k-r}}\cdot {{b}^{r}}$
Substituting $a=1$, $b=x$, $k=2n$and $r=n$, we get the following expression,
$\begin{aligned} & {{T}_{n+1}}={}^{2n}{{C}_{n}}\cdot {{(1)}^{2n-n}}\cdot {{x}^{n}} \\\ & =\dfrac{2n!}{n!\left( 2n-n \right)!}\cdot {{(1)}^{n}}\cdot {{x}^{n}} \\\ & =\dfrac{2n!}{n!n!}\cdot {{x}^{n}} \end{aligned}$
So, the middle term of ${{(1+x)}^{2n}}$ is ${{T}_{n+1}}=\dfrac{2n!}{n!n!}\cdot {{x}^{n}}$. Now, we will simplify the middle term in the following manner,
First we will expand the factorials in the numerator.
${{T}_{n+1}}=\dfrac{1\cdot 2\cdot \ldots \cdot (2n-1)\cdot (2n)}{n!n!}\cdot {{x}^{n}}$
We will separate the odd terms and the even terms in the numerator.
${{T}_{n+1}}=\dfrac{\left( 1\cdot 3\cdot \ldots \cdot (2n-1) \right)\cdot \left( 2\cdot 4\cdot \ldots \cdot 2n \right)}{n!n!}\cdot {{x}^{n}}$
The collected even terms in the numerator can be rewritten as multiples of 2 and then factor out all the 2's in the following manner,
$\begin{aligned} & {{T}_{n+1}}=\dfrac{\left( 1\cdot 3\cdot \ldots \cdot (2n-1) \right)\cdot \left( 2\cdot 1\times 2\cdot 2\times 3\cdot \ldots \cdot 2\times (n-1)\cdot 2\times n \right)}{n!n!}\cdot {{x}^{n}} \\\ & =\dfrac{\left( 1\cdot 3\cdot \ldots \cdot (2n-1) \right)\cdot \left( 1\cdot 2\cdot 3\cdot \ldots \cdot (n-1)\cdot n \right)\cdot {{2}^{n}}}{n!n!}\cdot {{x}^{n}} \\\ & =\dfrac{\left( 1\cdot 3\cdot \ldots \cdot (2n-1) \right)\cdot n!}{n!n!}\cdot {{2}^{n}}{{x}^{n}} \\\ & =\dfrac{\left( 1\cdot 3\cdot \ldots \cdot (2n-1) \right)}{n!}\cdot {{2}^{n}}{{x}^{n}} \end{aligned}$
Hence, proved.

Note: The formula for the general term of the expansion of ${{(a+b)}^{k}}$ is the main part of this question. If $k$ is odd, then the middle term is the ${{\left( \dfrac{n+1}{2} \right)}^{th}}$ term. We should be able to observe how to factorize the numerator of the middle term in order to get the desired expression.