Question

Find the coefficient of ${{x}^{n}}\text{ in }\dfrac{\left( 1+x \right)\left( 1+2x \right)\left( 1+3x \right)}{\left( 1-x \right)\left( 1-2x \right)\left( 1-3x \right)}.$
(a) $12-30\cdot {{2}^{n}}-20\cdot {{3}^{n}}$
(b) $12-30\cdot {{2}^{n}}+20\cdot {{3}^{n}}$
(c) $12+30\cdot {{2}^{n}}+20\cdot {{3}^{n}}$
(d) $12+30\cdot {{2}^{n}}-20\cdot {{3}^{n}}$

Answer 1

Hint: We must calculate the partial fraction of the denominator and then use the binomial theorem expansion to expand the expression. We then have to sort out the terms containing ${{x}^{n}}$ and simplify it further to get the required result.

Complete step-by-step solution:
We need to find the coefficient of ${{x}^{n}}$ , so it is clear that we need to use the Binomial theorem.
Let us first focus on the denominator term, and find its partial fraction.
$\dfrac{1}{\left( 1-x \right)\left( 1-2x \right)\left( 1-3x \right)}=\dfrac{A}{1-x}+\dfrac{B}{1-2x}+\dfrac{C}{1-3x}$
We now need to find the values of A, B and C.
So, let us solve the RHS part,
$\dfrac{1}{\left( 1-x \right)\left( 1-2x \right)\left( 1-3x \right)}=\dfrac{A\left( 1-5x+{{x}^{2}} \right)+B\left( 1-4x+3{{x}^{2}} \right)+C\left( 1-3x+2{{x}^{2}} \right)}{\left( 1-x \right)\left( 1-2x \right)\left( 1-3x \right)}$
We can rearrange the terms on RHS, to get
$\dfrac{1}{\left( 1-x \right)\left( 1-2x \right)\left( 1-3x \right)}=\dfrac{{{x}^{2}}\left( 6A+3B+2C \right)+x\left( -5A-4B-3C \right)+\left( A+B+C \right)}{\left( 1-x \right)\left( 1-2x \right)\left( 1-3x \right)}$
On comparing LHS and RHS, we get
$\begin{align}
  & 6A+3B+2C=0 \\
 & -5A-4B-3C=0 \\
 & A+B+C=1 \\
\end{align}$
On solving, these equations simultaneously, we get
$A=\dfrac{1}{2},B=-4,C=\dfrac{9}{2}$
Thus, we now have
$\dfrac{1}{\left( 1-x \right)\left( 1-2x \right)\left( 1-3x \right)}=\dfrac{1}{2\left( 1-x \right)}-\dfrac{4}{\left( 1-2x \right)}+\dfrac{9}{2\left( 1-3x \right)}$
Let us now include the numerator term,
$\dfrac{\left( 1+x \right)\left( 1+2x \right)\left( 1+3x \right)}{\left( 1-x \right)\left( 1-2x \right)\left( 1-3x \right)}=\left( 1+x \right)\left( 1+2x \right)\left( 1+3x \right)\left[ \dfrac{1}{2\left( 1-x \right)}-\dfrac{4}{\left( 1-2x \right)}+\dfrac{9}{2\left( 1-3x \right)} \right]$
On multiplying the factors, we get
$\dfrac{\left( 1+x \right)\left( 1+2x \right)\left( 1+3x \right)}{\left( 1-x \right)\left( 1-2x \right)\left( 1-3x \right)}=\left( 6{{x}^{3}}+11{{x}^{2}}+6x+1 \right)\left[ \dfrac{1}{2\left( 1-x \right)}-\dfrac{4}{\left( 1-2x \right)}+\dfrac{9}{2\left( 1-3x \right)} \right]$
From Binomial theorem, we know the following expansions
$\begin{align}
  & \dfrac{1}{1-x}=1+x+{{x}^{2}}+{{x}^{3}}+... \\
 & \dfrac{1}{1-2x}=1+2x+{{\left( 2x \right)}^{2}}+{{\left( 2x \right)}^{3}}+... \\
 & \dfrac{1}{1-3x}=1+3x+{{\left( 3x \right)}^{2}}+{{\left( 3x \right)}^{3}}+... \\
\end{align}$
Using these expansions in the RHS part of our equation, we get the following result
$\begin{align}
  & \dfrac{\left( 1+x \right)\left( 1+2x \right)\left( 1+3x \right)}{\left( 1-x \right)\left( 1-2x \right)\left( 1-3x \right)}= \\
 & \left( 6{{x}^{3}}+11{{x}^{2}}+6x+1 \right)\left[ \dfrac{1}{2}\left( 1+x+{{x}^{2}}+{{x}^{3}}+... \right)-4\left( 1+2x+{{\left( 2x \right)}^{2}}+{{\left( 2x \right)}^{3}}+... \right)+\dfrac{9}{2}\left( 1+3x+{{\left( 3x \right)}^{2}}+{{\left( 3x \right)}^{3}}+... \right) \right] \\
\end{align}$
We are only concerned about the coefficient of ${{x}^{n}}$ term. So, we can neglect all other terms and focus on just the term with ${{x}^{n}}$.
\[\begin{align}
  & \text{Terms with }{{x}^{n}}=\dfrac{1}{2}\left[ 6{{x}^{n}}+11{{x}^{n}}+6{{x}^{n}}+{{x}^{n}} \right]-4\left[ 6\cdot {{2}^{n-3}}{{x}^{n}}+11\cdot {{2}^{n-2}}{{x}^{n}}+6\cdot {{2}^{n-1}}{{x}^{n}}+{{2}^{n}}{{x}^{n}} \right] \\
 & +\dfrac{9}{2}\left[ 6\cdot {{3}^{n-3}}{{x}^{n}}+11\cdot {{3}^{n-2}}{{x}^{n}}+6\cdot {{3}^{n-1}}{{x}^{n}}+{{3}^{n}}{{x}^{n}} \right] \\
\end{align}\]
Hence, the coefficient of ${{x}^{n}}$ is = \[\dfrac{1}{2}\left[ 6+11+6+1 \right]-4\left[ 6\cdot {{2}^{n-3}}+11\cdot {{2}^{n-2}}+6\cdot {{2}^{n-1}}+{{2}^{n}} \right]+\dfrac{9}{2}\left[ 6\cdot {{3}^{n-3}}+11\cdot {{3}^{n-2}}+6\cdot {{3}^{n-1}}+{{3}^{n}} \right]\]
We can further solve the above equation as
Coefficient = \[12-24\cdot {{2}^{n-3}}-44\cdot {{2}^{n-2}}-24\cdot {{2}^{n-1}}-4\cdot {{2}^{n}}+27\cdot {{3}^{n-3}}+\dfrac{99}{2}\cdot {{3}^{n-2}}+27\cdot {{3}^{n-1}}+\dfrac{9}{2}{{3}^{n}}\]
This can further be reduced as
Coefficient = \[12-3\cdot {{2}^{n}}-11\cdot {{2}^{n}}-12\cdot {{2}^{n}}-4\cdot {{2}^{n}}+{{3}^{n}}+\dfrac{11}{2}\cdot {{3}^{n}}+9\cdot {{3}^{n}}+\dfrac{9}{2}{{3}^{n}}\]
Hence, we can now clearly write
Coefficient = $12-30\cdot {{2}^{n}}+20\cdot {{3}^{n}}$
Hence, option (b) is the correct answer.

Note: We must be very careful while solving this problem, as it has some very complicated terms, and has a high chance of calculation mistake. We must use the correct expansions for all terms. And we must not miss any term while calculating the coefficient.