Answer
Verified
424.5k+ 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
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
Write an application to the principal requesting five class 10 english CBSE
What is the type of food and mode of feeding of the class 11 biology CBSE
Name 10 Living and Non living things class 9 biology CBSE