AP DSC School Assistant Math Model Test 2

Concept:The problem involves finding the coefficient of xⁿ in the expansion of a rational function. We use partial fractions to split the expression into simpler terms, then apply the binomial series for $(1 - u)^{-1}$ which expands as $1 + u + u^2 + \dots + u^n + \dots$ (valid for $|u|<1$).Explanation:Step 1: Write the given expression and decompose it into partial fractions.$$\frac{1}{(1-x)(3-x)} = \frac{A}{1-x} + \frac{B}{3-x}$$Find $A$ and $B$ by solving: $A = \lim_{x\to 1}\frac{1}{3-x} = \frac{1}{2}$, and $B = \lim_{x\to 3}\frac{1}{1-x} = \frac{1}{-2} = -\frac{1}{2}$.Thus,$$\frac{1}{(1-x)(3-x)} = \frac{1}{2}\left(\frac{1}{1-x} - \frac{1}{3-x}\right).$$Step 2: Rewrite the second term to use the standard binomial form.$$\frac{1}{3-x} = \frac{1}{3}\cdot\frac{1}{1-\frac{x}{3}} = \frac{1}{3}\left(1-\frac{x}{3}\right)^{-1}.$$Step 3: Expand each $(1 - u)^{-1}$ using the binomial series:$$(1-x)^{-1} = 1 + x + x^2 + \dots + x^n + \dots$$$$\left(1-\frac{x}{3}\right)^{-1} = 1 + \frac{x}{3} + \left(\frac{x}{3}\right)^2 + \dots + \left(\frac{x}{3}\right)^n + \dots$$Step 4: The entire expression becomes:$$\frac{1}{2}\left[ (1-x)^{-1} - \frac{1}{3}\left(1-\frac{x}{3}\right)^{-1} \right].$$The coefficient of $x^n$ in $(1-x)^{-1}$ is $1$. The coefficient of $x^n$ in $\frac{1}{3}\left(1-\frac{x}{3}\right)^{-1}$ is $\frac{1}{3} \cdot \left(\frac{1}{3}\right)^n = \frac{1}{3^{n+1}}$.Step 5: Therefore, the coefficient of $x^n$ in the original expression is:$$\frac{1}{2}\left[ 1 - \frac{1}{3^{n+1}} \right] = \frac{1}{2}\cdot\frac{3^{n+1} - 1}{3^{n+1}} = \frac{3^{n+1} - 1}{2\cdot 3^{n+1}}.$$Answer:Option A: $\frac{3^{n+1}-1}{2\cdot 3^{n+1}}$