Question

The sum of the coefficients of the expression ${{\left( \dfrac{1}{x}+2x \right)}^{6}}$ is equal to
[a] 1024
[b] 729
[c] 243
[d] 512
[e] 64

Answer 1

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 ${{9}^{9}}$.