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Hint: First of all by using the formula for general terms of binomial expansion that is, \[Tn={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}\], write the general term of ${{\left( x+\dfrac{1}{x} \right)}^{4}}$by taking $a=x,b=\dfrac{1}{x}$ and n=4. Then put the power of x = 0 to find the term independent of x.

Complete step-by-step answer:

Let us consider the expression given in question as

$A={{\left( x+\dfrac{1}{x} \right)}^{4}}$…………… (1)

We know that, by binomial theorem, we can expand ${{\left( a+b \right)}^{n}}$ as,

${{\left( a+b \right)}^{n}}={}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{b}^{3}}+.......{}^{n}{{C}_{n}}{{a}^{1}}{{b}^{n-1}}$

We can also write it as,

${{\left( a+b \right)}^{n}}=\sum\limits_{r=1}^{n}{{}^{n}{{C}_{r}}}{{a}^{n-r}}{{b}^{r}}$

Therefore, we get general term in expansion of

${{\left( a+b \right)}^{n}}\ as\ {}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$

By taking $a=x,b=\dfrac{1}{x}\And n=4$we get general term of ${{\left( x+\dfrac{1}{x} \right)}^{4}}$as,

General term of ${{\left( x+\dfrac{1}{x} \right)}^{4}}={}^{4}{{C}_{r}}{{\left( x \right)}^{4-r}}{{\left( \dfrac{1}{x} \right)}^{r}}$

Let us consider this general term as,

$Tn={}^{4}{{C}_{r}}{{\left( x \right)}^{4-r}}{{\left[ \dfrac{1}{x} \right]}^{r}}$

We know that ${{\left( \dfrac{a}{b} \right)}^{n}}=\dfrac{{{a}^{n}}}{{{b}^{n}}}$ , applying this in above expression, we get,

$Tn={}^{4}{{C}_{r}}{{\left( x \right)}^{4-r}}\dfrac{{{\left( 1 \right)}^{r}}}{{{\left( x \right)}^{r}}}$

Or we get $Tn={}^{4}{{C}_{r}}{{\left( x \right)}^{4-r}}\dfrac{1}{{{\left( x \right)}^{r}}}$

We can also write above expression as,

\[Tn=\dfrac{{}^{4}{{C}_{r}}{{\left( x \right)}^{4-r}}}{{{\left( x \right)}^{r}}}\]

Now, we know that $\dfrac{{{a}^{p}}}{{{a}^{q}}}={{a}^{p-q}}$. By applying this in above expression, we get

\[Tn={}^{4}{{C}_{r}}{{\left( x \right)}^{\left( 4-r \right)-r}}\]

Therefore, we get \[Tn={}^{4}{{C}_{r}}{{\left( x \right)}^{4-2r}}\]……………. (2)

Now, to find the term which is independent of x, we must put the power of x = 0.

Therefore we get, 4-2r=0

By taking the terms containing ‘r’ to one side and constant term to other side, we get,

$\Rightarrow 2r=4$

By dividing 2 on both sides, we get,

$\begin{align}

& \Rightarrow \dfrac{2r}{2}=\dfrac{4}{2} \\

& \Rightarrow r=2 \\

\end{align}$

Now to get the term independent of x, we put r= 2 in equation (2), we get,

$\begin{align}

& Tn={}^{4}{{C}_{2}}{{\left( x \right)}^{4-2\left( 2 \right)}} \\

& Tn={}^{4}{{C}_{2}}{{x}^{4-4}} \\

& Tn={}^{4}{{C}_{2}}{{\left( x \right)}^{0}} \\

\end{align}$

We know that ${{\left( a \right)}^{0}}=1$, therefore we get.

$Tn={}^{4}{{C}_{2}}$

$\begin{align}

& Tn=\dfrac{4!}{2!2!} \\

& Tn=6 \\

\end{align}$

Therefore, we get the term independent of x in ${{\left( x+\dfrac{1}{x} \right)}^{4}}$as ${}^{4}{{C}_{2}}=6$.

Note: Students must note that when they are asked to find the term independent of variable, they should always put the power of that variable = 0. Also students must take special care while writing each term and cross check if they have written it correctly or not. Students often make mistakes while writing the powers and this must be avoided.

Complete step-by-step answer:

Let us consider the expression given in question as

$A={{\left( x+\dfrac{1}{x} \right)}^{4}}$…………… (1)

We know that, by binomial theorem, we can expand ${{\left( a+b \right)}^{n}}$ as,

${{\left( a+b \right)}^{n}}={}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{b}^{3}}+.......{}^{n}{{C}_{n}}{{a}^{1}}{{b}^{n-1}}$

We can also write it as,

${{\left( a+b \right)}^{n}}=\sum\limits_{r=1}^{n}{{}^{n}{{C}_{r}}}{{a}^{n-r}}{{b}^{r}}$

Therefore, we get general term in expansion of

${{\left( a+b \right)}^{n}}\ as\ {}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$

By taking $a=x,b=\dfrac{1}{x}\And n=4$we get general term of ${{\left( x+\dfrac{1}{x} \right)}^{4}}$as,

General term of ${{\left( x+\dfrac{1}{x} \right)}^{4}}={}^{4}{{C}_{r}}{{\left( x \right)}^{4-r}}{{\left( \dfrac{1}{x} \right)}^{r}}$

Let us consider this general term as,

$Tn={}^{4}{{C}_{r}}{{\left( x \right)}^{4-r}}{{\left[ \dfrac{1}{x} \right]}^{r}}$

We know that ${{\left( \dfrac{a}{b} \right)}^{n}}=\dfrac{{{a}^{n}}}{{{b}^{n}}}$ , applying this in above expression, we get,

$Tn={}^{4}{{C}_{r}}{{\left( x \right)}^{4-r}}\dfrac{{{\left( 1 \right)}^{r}}}{{{\left( x \right)}^{r}}}$

Or we get $Tn={}^{4}{{C}_{r}}{{\left( x \right)}^{4-r}}\dfrac{1}{{{\left( x \right)}^{r}}}$

We can also write above expression as,

\[Tn=\dfrac{{}^{4}{{C}_{r}}{{\left( x \right)}^{4-r}}}{{{\left( x \right)}^{r}}}\]

Now, we know that $\dfrac{{{a}^{p}}}{{{a}^{q}}}={{a}^{p-q}}$. By applying this in above expression, we get

\[Tn={}^{4}{{C}_{r}}{{\left( x \right)}^{\left( 4-r \right)-r}}\]

Therefore, we get \[Tn={}^{4}{{C}_{r}}{{\left( x \right)}^{4-2r}}\]……………. (2)

Now, to find the term which is independent of x, we must put the power of x = 0.

Therefore we get, 4-2r=0

By taking the terms containing ‘r’ to one side and constant term to other side, we get,

$\Rightarrow 2r=4$

By dividing 2 on both sides, we get,

$\begin{align}

& \Rightarrow \dfrac{2r}{2}=\dfrac{4}{2} \\

& \Rightarrow r=2 \\

\end{align}$

Now to get the term independent of x, we put r= 2 in equation (2), we get,

$\begin{align}

& Tn={}^{4}{{C}_{2}}{{\left( x \right)}^{4-2\left( 2 \right)}} \\

& Tn={}^{4}{{C}_{2}}{{x}^{4-4}} \\

& Tn={}^{4}{{C}_{2}}{{\left( x \right)}^{0}} \\

\end{align}$

We know that ${{\left( a \right)}^{0}}=1$, therefore we get.

$Tn={}^{4}{{C}_{2}}$

$\begin{align}

& Tn=\dfrac{4!}{2!2!} \\

& Tn=6 \\

\end{align}$

Therefore, we get the term independent of x in ${{\left( x+\dfrac{1}{x} \right)}^{4}}$as ${}^{4}{{C}_{2}}=6$.

Note: Students must note that when they are asked to find the term independent of variable, they should always put the power of that variable = 0. Also students must take special care while writing each term and cross check if they have written it correctly or not. Students often make mistakes while writing the powers and this must be avoided.

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