Answer
451.8k+ views
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)