Question

The coefficient of $x^4$ in the expansion of ${\{\sqrt{1+x^2}-x\}}^{-1}$ in ascending powers of $x$, when $\left|x\right|<1$, is

• A. $0$
• B. $\frac{1}{2}$
• C. $-\frac{1}{2}$
• D. $-\frac{1}{8}$

Answer 1

Answer: (D)

$-\frac{1}{8}$

$\frac{1}{\sqrt{1+x^2}-x}$
$=\frac{x+\sqrt{1+x^2}}{1+x^2-x^2}$
$=x+\sqrt{1+x^2}=x+(1+\frac{1}{2}{x}^2+\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}{x}^4...\infty )$
Hence coefficient of $x^4$ is
$\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}=\frac{-1}{4(2!)}=\frac{-1}{8}$