Question

Coefficient of $x^n$ in $\frac{(1+x)(1+2x)(1+3x)}{(1-x)(1-2x)(1-3x)}$ is

• A. $12-{30.2}^n-{20.3}^n$
• B. $12-{30.2}^n+{20.3}^n$
• C. $12+{30.2}^n+{20.3}^n$
• D. $12+{30.2}^n-{20.3}^n$

Answer 1

The correct option is B $12-{30.2}^n+{20.3}^n$

First we write the expression as sum of three terms using partial fraction.
To use partial fraction, first wirte the given expression as $\frac{(1+x)(1+2x)(1+3x)}{(1-x)(1-2x)(1-3x)}=\frac{Ax+B}{1-x}+\frac{C}{1-2x}+\frac{D}{1-3x}$
Now, simplpifying and comparing the cooefficients of $x^3$ on both sides, we get
$6=6A\Rightarrow A=1$
Similarly comparing the coefficients of $x^2$ on both sides, we get
$11=-2A-3A+6B+3C+2D$
$\Rightarrow 6B+3C+2D=16\dots \left(1\right)$ (using $A=1$)
Comparing the coefficients of $x$ on both sides, we get
$6=A-2B-3B-4C-3D$
$\Rightarrow -5B-4C-3D=5\dots \left(2\right)$
Finally, comparing the constant terms, we get
$B+C+D=1\dots \left(3\right)$
Solving $\left(1\right),\left(2\right),\left(3\right)$ simultaneously, we get $B=11,C=-30,D=20$
Hence, the given expression can be written as
$\frac{x+11}{1-x}+\frac{-30}{1-2x}+\frac{20}{1-3x}$
$=(x+11){(1-x)}^{-1}+(-30){(1-2x)}^{-1}+20{(1-3x)}^{-1}$
$=(x+11)(1+x+x^2+x^3+...+x^{n-1}+x^n+...)+(-30)(1+2x+2^2{x}^2+...2^n{x}^n+...)+20(1+3x+3^2{x}^2+...+3^n{x}^n+...)$
Hence, the coefficient of $x^n$ is equal to $12-30.2^n+20.3^n$