Question

The coefficient of $x^5$ in the expansion of $\frac{1+x^2}{a+x},|x|<1$, is

• A. $(-1)[{\left(\frac{1}{a}\right)}^6+{\left(\frac{1}{a}\right)}^4]$
• B. $(-1)[\left(\frac{1}{a}\right)+{\left(\frac{1}{a}\right)}^4]$
• C. 0
• D. -2

Answer 1

Answer: (A)

$(-1)[{\left(\frac{1}{a}\right)}^6+{\left(\frac{1}{a}\right)}^4]$

Let $S=\frac{1+x^2}{a+x},|x|<1$

$S=(1+x^2)(a+x{)}^{-1}$

$S=a^{-1}(1+x^2)(1+\frac{x}{a}{)}^{-1}$

$(1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3...$

Put $n=-1$ and $x=\frac{x}{a}$:

${(1+\frac{x}{a})}^{-1}=1-\left(\frac{x}{a}\right)+{\left(\frac{x}{a}\right)}^2-{\left(\frac{x}{a}\right)}^3...=\sum _0^{\infty }{\left(\frac{-1}{a}\right)}^n{x}^n$

Substituting in $S$:

$S=a^{-1}(1+x^2)(1-(\frac{x}{a})+(\frac{x}{a}{)}^2-(\frac{x}{a}{)}^3...)$

$\therefore$ $S=a^{-1}(1-\left(\frac{x}{a}\right)+\sum _2^{\infty }[{\left(\frac{-1}{a}\right)}^n+{\left(\frac{-1}{a}\right)}^{n-2}]x^n)$

The coefficient of $x^5$ (Say $\beta$) in $S=a^{-1}[{\left(\frac{-1}{a}\right)}^5+{\left(\frac{-1}{a}\right)}^3]$

$\therefore \beta =(-1)[{\left(\frac{1}{a}\right)}^6+{\left(\frac{1}{a}\right)}^4]$