Question

If $x>\sqrt{3}$ and $\frac{x^2+1}{(x^2+2)(x^2+3)}$ is expanded in terms of powers of $x$, then the coefficient of $x^{-8}$ is

• A. \( 0 \)
• B. \( -81 \)
• C. \( 46 \)
• D. \( -46 \)

Hint:

When expanding rational functions in powers of $x$ (especially negative powers), first perform partial fraction decomposition. For terms like $\frac{1}{ax^k+b}$, factor out $ax^k$ to get $\frac{1}{ax^k}\left(1+\frac{b}{ax^k}\right)^{-1}$. The condition for $x$ (e.g., $x>\sqrt{3}$ here) is crucial as it determines which term to factor out to ensure the absolute value of the ratio is less than 1, making the binomial series converge. For $(1+u)^{-1}$, the general term is $(-1)^n u^n$. For $(1-u)^{-1}$, it's $u^n$.

Answer 1

The Correct Option is D

Let the given expression be $E = \frac{x^2+1}{(x^2+2)(x^2+3)}$. We first decompose the rational function into partial fractions. Let $y = x^2$. Then $E = \frac{y+1}{(y+2)(y+3)}$. We set up the partial fraction decomposition as: $\frac{y+1}{(y+2)(y+3)} = \frac{A}{y+2} + \frac{B}{y+3}$ To find $A$ and $B$, multiply both sides by $(y+2)(y+3)$: $y+1 = A(y+3) + B(y+2)$ Set $y = -2$: $-2+1 = A(-2+3) + B(-2+2)$ $-1 = A(1) + B(0)$ $A = -1$ Set $y = -3$: $-3+1 = A(-3+3) + B(-3+2)$ $-2 = A(0) + B(-1)$ $-2 = -B$ $B = 2$ So, the expression can be rewritten as: $E = \frac{-1}{x^2+2} + \frac{2}{x^2+3}$ Now, we need to expand each term in powers of $x$. Since $x>\sqrt{3}$, we have $x^2>3$. This means $\frac{2}{x^2}<1$ and $\frac{3}{x^2}<1$, which allows us to use the binomial series expansion of $(1+u)^{-1}$. Recall the binomial expansion: $(1+u)^{-1} = 1 - u + u^2 - u^3 + u^4 - \dots + (-1)^k u^k + \dots$ Consider the first term: $\frac{-1}{x^2+2}$ $\frac{-1}{x^2+2} = -1(x^2+2)^{-1} = -1 \left[x^2\left(1+\frac{2}{x^2}\right)\right]^{-1}$ $= -1 \cdot (x^2)^{-1} \left(1+\frac{2}{x^2}\right)^{-1}$ $= -x^{-2} \left(1 - \left(\frac{2}{x^2}\right) + \left(\frac{2}{x^2}\right)^2 - \left(\frac{2}{x^2}\right)^3 + \left(\frac{2}{x^2}\right)^4 - \dots\right)$ $= -x^{-2} \left(1 - 2x^{-2} + 4x^{-4} - 8x^{-6} + 16x^{-8} - \dots\right)$ $= -x^{-2} + 2x^{-4} - 4x^{-6} + 8x^{-8} - 16x^{-10} + \dots$ The coefficient of $x^{-8}$ from this term is $8$. Consider the second term: $\frac{2}{x^2+3}$ $\frac{2}{x^2+3} = 2(x^2+3)^{-1} = 2 \left[x^2\left(1+\frac{3}{x^2}\right)\right]^{-1}$ $= 2 \cdot (x^2)^{-1} \left(1+\frac{3}{x^2}\right)^{-1}$ $= 2x^{-2} \left(1 - \left(\frac{3}{x^2}\right) + \left(\frac{3}{x^2}\right)^2 - \left(\frac{3}{x^2}\right)^3 + \left(\frac{3}{x^2}\right)^4 - \dots\right)$ $= 2x^{-2} \left(1 - 3x^{-2} + 9x^{-4} - 27x^{-6} + 81x^{-8} - \dots\right)$ $= 2x^{-2} - 6x^{-4} + 18x^{-6} - 54x^{-8} + 162x^{-10} - \dots$ The coefficient of $x^{-8}$ from this term is $-54$. To find the total coefficient of $x^{-8}$ in the expansion of $E$, we sum the coefficients from both terms: Total coefficient of $x^{-8} = 8 + (-54) = 8 - 54 = -46$.