Question

The absolute value of the difference of the coefficients of $x^4$ and $x^6$ in the expansion of  
$\frac{2x^2}{(x^2+1)(x^2+2)}$ 
is:

• A. \( \frac{13}{4} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{9}{4} \)
• D. \( 1 \)

Hint:

Use partial fraction decomposition to simplify rational functions before applying binomial expansion for coefficient extraction.

Answer 1

The Correct Option is A

Step 1: Partial Fraction Expansion
Expanding the function: \[ \frac{2x^2}{(x+1)(x+2)} = A(x+2) + B(x+1) \] Solving for \( A \) and \( B \), we find: \[ A = 1, \quad B = -\frac{1}{2} \] Step 2: Expanding Each Term
Using binomial series, we expand: \[ \frac{1}{(x+1)} = 1 - x + x^2 - x^3 + \dots \] \[ \frac{1}{(x+2)} = \frac{1}{2} (1 - x/2 + x^2/4 - x^3/8 + \dots) \] Multiplying by \( 2x^2 \), we extract coefficients of \( x^4 \) and \( x^6 \): \[ \text{Coefficient of } x^4 = \frac{5}{4}, \quad \text{Coefficient of } x^6 = -\frac{8}{4} \] Step 3: Finding Absolute Difference
\[ \left| \frac{5}{4} - \left(-\frac{8}{4} \right) \right| = \frac{13}{4} \] Thus, the correct answer is \( \frac{13}{4} \).