Answer
Verified
64.8k+ views
Hint: Expand the given expression by using binomial theorem and find the coefficients of the terms involved in the expression. Calculate the sum of these coefficients to get the result. Instead of making use of binomial theorem, you can also use Pascal's triangle to get the binomial coefficients.
Complete step-by-step answer:
We know from binomial theorem;
${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+\ldots +{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$
Putting n = 6, $x=\dfrac{1}{x}$ and $y=2x$ in the above expression, we get
${{\left( \dfrac{1}{x}+2x \right)}^{6}}={}^{6}{{C}_{0}}{{\left( \dfrac{1}{x} \right)}^{6}}{{\left( 2x \right)}^{0}}+{}^{6}{{C}_{1}}{{\left( \dfrac{1}{x} \right)}^{5}}{{\left( 2x \right)}^{1}}+{}^{6}{{C}_{2}}{{\left( \dfrac{1}{x} \right)}^{4}}{{\left( 2x \right)}^{2}}+{}^{6}{{C}_{3}}{{\left( \dfrac{1}{x} \right)}^{3}}{{\left( 2x \right)}^{3}}+{}^{6}{{C}_{4}}{{\left( \dfrac{1}{x} \right)}^{2}}{{\left( 2x \right)}^{4}}+{}^{6}{{C}_{5}}{{\left( \dfrac{1}{x} \right)}^{1}}{{\left( 2x \right)}^{5}}+{}^{6}{{C}_{6}}{{\left( \dfrac{1}{x} \right)}^{0}}{{\left( 2x \right)}^{6}}$
Now we know that \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Using the above formula, we get
$\begin{align}
& {}^{6}{{C}_{0}}=\dfrac{6!}{0!\left( 6-0 \right)!}=\dfrac{6!}{6!}=1 \\
& {}^{6}{{C}_{1}}=\dfrac{6!}{1!\left( 6-1 \right)!}=\dfrac{6\times 5!}{1!5!}=6 \\
& {}^{6}{{C}_{2}}=\dfrac{6!}{2!\left( 6-2 \right)!}=\dfrac{6\times 5\times 4!}{2!4!}=\dfrac{30}{2}=15 \\
& {}^{6}{{C}_{3}}=\dfrac{6!}{3!3!}=\dfrac{6\times 5\times 4\times 3!}{3!3!}=\dfrac{6\times 5\times 4}{6}=20 \\
\end{align}$
We know that ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$
Using we get
$\begin{align}
& {}^{6}{{C}_{4}}={}^{6}{{C}_{2}}=15 \\
& {}^{6}{{C}_{5}}={}^{6}{{C}_{1}}=6 \\
& {}^{6}{{C}_{6}}={}^{6}{{C}_{0}}=1 \\
\end{align}$
Hence we have
${{\left( \dfrac{1}{x}+2x \right)}^{6}}=1{{\left( \dfrac{1}{x} \right)}^{6}}{{\left( 2x \right)}^{0}}+6{{\left( \dfrac{1}{x} \right)}^{5}}{{\left( 2x \right)}^{1}}+15{{\left( \dfrac{1}{x} \right)}^{4}}{{\left( 2x \right)}^{2}}+20{{\left( \dfrac{1}{x} \right)}^{3}}{{\left( 2x \right)}^{3}}+15{{\left( \dfrac{1}{x} \right)}^{2}}{{\left( 2x \right)}^{4}}+6{{\left( \dfrac{1}{x} \right)}^{1}}{{\left( 2x \right)}^{5}}+1{{\left( \dfrac{1}{x} \right)}^{0}}{{\left( 2x \right)}^{6}}$
Simplifying, we get
${{\left( \dfrac{1}{x}+2x \right)}^{6}}=\dfrac{1}{{{x}^{6}}}+\dfrac{12}{{{x}^{4}}}+\dfrac{60}{{{x}^{2}}}+160+240{{x}^{2}}+192{{x}^{4}}+64{{x}^{6}}$
Hence the sum of coefficients = 1+12+60+160+240+192+64=729
Note: [1] Alternative solution 1: Construct pascal triangle till n = 6
$\begin{align}
& 1 \\
& 1\text{ 1} \\
& \text{1 2 1} \\
& \text{1 3 3 1} \\
& \text{1 4 6 4 1} \\
& \text{1 5 10 10 5 1} \\
& \text{1 6 15 20 15 6 1} \\
\end{align}$
Hence we have the binomial coefficients as 1, 6, 15, 20, 15, 6 and 1, which is the same as above.
[2] Alternative solution 2: Best Method.
Let the expansion of the given expression be ${{a}_{0}}{{x}^{6}}+{{a}_{1}}{{x}^{5}}+\ldots +{{a}_{12}}{{x}^{-6}}$
Hence we have
${{\left( \dfrac{1}{x}+2x \right)}^{6}}={{a}_{0}}{{x}^{6}}+{{a}_{1}}{{x}^{5}}+\ldots +{{a}_{12}}{{x}^{-6}}$
Put x = 1, we get
$\begin{align}
& {{\left( \dfrac{1}{1}+2\left( 1 \right) \right)}^{6}}={{a}_{0}}{{1}^{6}}+{{a}_{1}}{{1}^{5}}+\ldots +{{a}_{12}}{{1}^{-6}} \\
& \Rightarrow \sum\limits_{i=0}^{12}{{{a}_{i}}={{3}^{6}}=729} \\
\end{align}$
Hence the sum of the coefficients = 729.
[3] Some times, the question asks to find the value of the constant term. In that case, if $x=0$ is within the domain of the expression then put x = 0 to get the result, e.g. Find the constant term in the expansion of the expression ${{\left( 2{{x}^{3}}+3{{x}^{2}}+9 \right)}^{9}}$.
Since x = 0 is in the domain put x = 0 we get ${{\left( 2\times 0+3\times 0+9 \right)}^{9}}={{9}^{9}}$
Hence the constant term in the expansion of the expression is ${{9}^{9}}$.
Complete step-by-step answer:
We know from binomial theorem;
${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+\ldots +{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$
Putting n = 6, $x=\dfrac{1}{x}$ and $y=2x$ in the above expression, we get
${{\left( \dfrac{1}{x}+2x \right)}^{6}}={}^{6}{{C}_{0}}{{\left( \dfrac{1}{x} \right)}^{6}}{{\left( 2x \right)}^{0}}+{}^{6}{{C}_{1}}{{\left( \dfrac{1}{x} \right)}^{5}}{{\left( 2x \right)}^{1}}+{}^{6}{{C}_{2}}{{\left( \dfrac{1}{x} \right)}^{4}}{{\left( 2x \right)}^{2}}+{}^{6}{{C}_{3}}{{\left( \dfrac{1}{x} \right)}^{3}}{{\left( 2x \right)}^{3}}+{}^{6}{{C}_{4}}{{\left( \dfrac{1}{x} \right)}^{2}}{{\left( 2x \right)}^{4}}+{}^{6}{{C}_{5}}{{\left( \dfrac{1}{x} \right)}^{1}}{{\left( 2x \right)}^{5}}+{}^{6}{{C}_{6}}{{\left( \dfrac{1}{x} \right)}^{0}}{{\left( 2x \right)}^{6}}$
Now we know that \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Using the above formula, we get
$\begin{align}
& {}^{6}{{C}_{0}}=\dfrac{6!}{0!\left( 6-0 \right)!}=\dfrac{6!}{6!}=1 \\
& {}^{6}{{C}_{1}}=\dfrac{6!}{1!\left( 6-1 \right)!}=\dfrac{6\times 5!}{1!5!}=6 \\
& {}^{6}{{C}_{2}}=\dfrac{6!}{2!\left( 6-2 \right)!}=\dfrac{6\times 5\times 4!}{2!4!}=\dfrac{30}{2}=15 \\
& {}^{6}{{C}_{3}}=\dfrac{6!}{3!3!}=\dfrac{6\times 5\times 4\times 3!}{3!3!}=\dfrac{6\times 5\times 4}{6}=20 \\
\end{align}$
We know that ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$
Using we get
$\begin{align}
& {}^{6}{{C}_{4}}={}^{6}{{C}_{2}}=15 \\
& {}^{6}{{C}_{5}}={}^{6}{{C}_{1}}=6 \\
& {}^{6}{{C}_{6}}={}^{6}{{C}_{0}}=1 \\
\end{align}$
Hence we have
${{\left( \dfrac{1}{x}+2x \right)}^{6}}=1{{\left( \dfrac{1}{x} \right)}^{6}}{{\left( 2x \right)}^{0}}+6{{\left( \dfrac{1}{x} \right)}^{5}}{{\left( 2x \right)}^{1}}+15{{\left( \dfrac{1}{x} \right)}^{4}}{{\left( 2x \right)}^{2}}+20{{\left( \dfrac{1}{x} \right)}^{3}}{{\left( 2x \right)}^{3}}+15{{\left( \dfrac{1}{x} \right)}^{2}}{{\left( 2x \right)}^{4}}+6{{\left( \dfrac{1}{x} \right)}^{1}}{{\left( 2x \right)}^{5}}+1{{\left( \dfrac{1}{x} \right)}^{0}}{{\left( 2x \right)}^{6}}$
Simplifying, we get
${{\left( \dfrac{1}{x}+2x \right)}^{6}}=\dfrac{1}{{{x}^{6}}}+\dfrac{12}{{{x}^{4}}}+\dfrac{60}{{{x}^{2}}}+160+240{{x}^{2}}+192{{x}^{4}}+64{{x}^{6}}$
Hence the sum of coefficients = 1+12+60+160+240+192+64=729
Note: [1] Alternative solution 1: Construct pascal triangle till n = 6
$\begin{align}
& 1 \\
& 1\text{ 1} \\
& \text{1 2 1} \\
& \text{1 3 3 1} \\
& \text{1 4 6 4 1} \\
& \text{1 5 10 10 5 1} \\
& \text{1 6 15 20 15 6 1} \\
\end{align}$
Hence we have the binomial coefficients as 1, 6, 15, 20, 15, 6 and 1, which is the same as above.
[2] Alternative solution 2: Best Method.
Let the expansion of the given expression be ${{a}_{0}}{{x}^{6}}+{{a}_{1}}{{x}^{5}}+\ldots +{{a}_{12}}{{x}^{-6}}$
Hence we have
${{\left( \dfrac{1}{x}+2x \right)}^{6}}={{a}_{0}}{{x}^{6}}+{{a}_{1}}{{x}^{5}}+\ldots +{{a}_{12}}{{x}^{-6}}$
Put x = 1, we get
$\begin{align}
& {{\left( \dfrac{1}{1}+2\left( 1 \right) \right)}^{6}}={{a}_{0}}{{1}^{6}}+{{a}_{1}}{{1}^{5}}+\ldots +{{a}_{12}}{{1}^{-6}} \\
& \Rightarrow \sum\limits_{i=0}^{12}{{{a}_{i}}={{3}^{6}}=729} \\
\end{align}$
Hence the sum of the coefficients = 729.
[3] Some times, the question asks to find the value of the constant term. In that case, if $x=0$ is within the domain of the expression then put x = 0 to get the result, e.g. Find the constant term in the expansion of the expression ${{\left( 2{{x}^{3}}+3{{x}^{2}}+9 \right)}^{9}}$.
Since x = 0 is in the domain put x = 0 we get ${{\left( 2\times 0+3\times 0+9 \right)}^{9}}={{9}^{9}}$
Hence the constant term in the expansion of the expression is ${{9}^{9}}$.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
According to classical free electron theory A There class 11 physics JEE_Main
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main