Question

If $\left|x\right|<1$, the coefficient of $x^n$ in the expansion of $\frac{1}{x^2-3x+2}$ is

• A. $1-\frac{1}{2^n}$
• B. $1-\frac{1}{2^{n+1}}$
• C. $1-2^n$
• D. none of these

Answer 1

Answer: (B)

$1-\frac{1}{2^{n+1}}$

$\frac{1}{(x-1)(x-2)}=\frac{1}{2}\left[\right(1-x{)}^{-1}(1-\frac{x}{2}{)}^{-1}]$
$=\frac{1}{2}\left[\right(1+x+x^2+x^3+...\infty \left)\right(1+\frac{x}{2}+\frac{x^2}{4}+...\infty \left)\right]$
$=p_0+p_{1}x+p_2\left(x^2\right)...\infty$
Hence from the above expansion, we get
$P_0=\frac{1}{2}$
$P_1=\frac{1}{2}(1+\frac{1}{2})=\frac{3}{4}$
$P_2=\frac{1}{2}(\frac{1}{4}+\frac{1}{2}+1)=\frac{7}{8}$
Hence coefficient of $P_n$
$=\frac{1}{2}[1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...\frac{1}{2^n}]$
$=\frac{1}{2}\left(\frac{1-\frac{1}{2^{n+1}}}{1-\frac{1}{2}}\right)$
$=\frac{1}{2}\left[2\right(1-\frac{1}{2^{n+1}}\left)\right]$
$=1-\frac{1}{2^{n+1}}$