Question

Prove that the term independent of x in the expansion of ${{\left( x+\dfrac{1}{x} \right)}^{2n}}\text{ }$ is $\dfrac{1.3.5...\left( 2n-1 \right)}{n!}{{.2}^{n}}$.

Answer 1

Hint: To prove that the term independent of x in the expansion of ${{\left( x+\dfrac{1}{x} \right)}^{2n}}\text{ }$ is $\dfrac{1.3.5...\left( 2n-1 \right)}{n!}{{.2}^{n}}$ , we will use the formula ${{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}.{{b}^{r}}$ for the expansion of ${{\left( x+\dfrac{1}{x} \right)}^{2n}}\text{ }$ which is for the form ${{\left( a+b \right)}^{n}}$ . We will get ${{T}_{r+1}}{{=}^{2n}}{{C}_{r}}{{\left( x \right)}^{2n-2r}}$ . To become independent of x, the exponent $2n-2r$ must be 0. Hence, $n=r$ . We can write the coefficient of term independent of x by substituting $n=r$ in $^{2n}{{C}_{r}}$ . We will get $^{2n}{{C}_{n}}$ . Now expand this using $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ to get $\dfrac{1\times 2\times 3\times 4\times 5\times 6\times ...\times \left( 2n-1 \right)\times 2n}{n!n!}$ . By rearranging, we get \[\dfrac{\left[ 1\times 3\times 5\times 7\times ...\times \left( 2n-1 \right) \right]\times {{2}^{n}}\left[ 1\times 2\times 3\times ...\times n \right]}{n!n!}\] . By solving further, we will reach the required result.

Complete step-by-step solution:
We have to prove that the term independent of x in the expansion of ${{\left( x+\dfrac{1}{x} \right)}^{2n}}\text{ }$ is $\dfrac{1.3.5...\left( 2n-1 \right)}{n!}{{.2}^{n}}$ .
Let us consider ${{\left( x+\dfrac{1}{x} \right)}^{2n}}\text{ }$ .
We know that the general term of expansion ${{\left( a+b \right)}^{n}}$ is given as
${{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}.{{b}^{r}}$
Hence, we can write the term of expansion ${{\left( x+\dfrac{1}{x} \right)}^{2n}}$ as
\[{{T}_{r+1}}{{=}^{2n}}{{C}_{r}}{{\left( x \right)}^{2n-r}}.{{\left( \dfrac{1}{x} \right)}^{r}}\]
We can write this as
\[{{T}_{r+1}}{{=}^{2n}}{{C}_{r}}{{\left( x \right)}^{2n-r}}.{{\left( x \right)}^{-r}}\]
We know that ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ .
$\begin{align}
  & \Rightarrow {{T}_{r+1}}{{=}^{2n}}{{C}_{r}}{{\left( x \right)}^{2n-r-r}} \\
 & \Rightarrow {{T}_{r+1}}{{=}^{2n}}{{C}_{r}}{{\left( x \right)}^{2n-2r}} \\
\end{align}$
To become independent of x, the exponent $2n-2r$ must be 0. Thus, we can write
$2n-2r=0$
When we solve this, we get
$\begin{align}
  & 2n=2r \\
 & \Rightarrow n=r \\
\end{align}$
Now, we can write the coefficient of term independent of x by substituting $n=r$ in $^{2n}{{C}_{r}}$ .
${{\Rightarrow }^{2n}}{{C}_{n}}$
We know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ .
$\begin{align}
  & {{\Rightarrow }^{2n}}{{C}_{n}}=\dfrac{2n!}{n!\left( 2n-n \right)!} \\
 & {{\Rightarrow }^{2n}}{{C}_{n}}=\dfrac{2n!}{n!n!} \\
\end{align}$
Let us expand the factorial. We know that $n!=1\times 2\times 3\times ...\times \left( n-1 \right)\times n$ . Hence,
$\dfrac{2n!}{n!n!}=\dfrac{1\times 2\times 3\times 4\times 5\times 6\times ...\times \left( 2n-1 \right)\times 2n}{n!n!}$
We can write this as follows by collecting odd and even terms separately.
\[\Rightarrow \dfrac{\left[ 1\times 3\times 5\times 7\times ...\times \left( 2n-1 \right) \right]\times \left[ 2\times 4\times 6\times ...\times 2n \right]}{n!n!}\]
Let us now take 2 commons from the even terms. We will get
\[\Rightarrow \dfrac{\left[ 1\times 3\times 5\times 7\times ...\times \left( 2n-1 \right) \right]\times {{2}^{n}}\left[ 1\times 2\times 3\times ...\times n \right]}{n!n!}\]
We know that \[n!=1\times 2\times 3\times ...\times n\] . Hence,
\[\Rightarrow \dfrac{\left[ 1\times 3\times 5\times 7\times ...\times \left( 2n-1 \right) \right]\times {{2}^{n}}n!}{n!n!}\]
Now, we can cancel n! from numerator and denominator. We will get
\[\Rightarrow \dfrac{\left[ 1\times 3\times 5\times 7\times ...\times \left( 2n-1 \right) \right]\times {{2}^{n}}}{n!}\]
Hence, the term independent of x in the expansion of ${{\left( x+\dfrac{1}{x} \right)}^{2n}}\text{ }$ is $\dfrac{1.3.5...\left( 2n-1 \right)}{n!}{{.2}^{n}}$ .
Thus proved.

Note: You may make mistake when writing the formula ${{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}.{{b}^{r}}$ as ${{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{\left( b \right)}^{n-r}}.{{a}^{r}}$ . Also, an error can be expected when writing $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ as $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n+r \right)!}$ . When taking 2 outside from the even set, be careful to write ${{2}^{n}}$ instead of 2.