Question

In the expansion of $\frac{e^x-1-x}{x^2}$ in ascending powers of $x$, the fourth term is

• A. $\frac{1}{5!}{x}^3$
• B. $\frac{1}{4!}{x}^4$
• C. $\frac{1}{3!}{x}^3$
• D. none of these

Answer 1

Answer: (A)

$\frac{1}{5!}{x}^3$

We know that
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\infty$
Hence $e^x-1-x$
$=\frac{x^2}{2!}+\frac{x^3}{3!}+...\infty$
Dividing by $x^2$ we get
$\frac{1}{2}+\frac{x}{3!}+\frac{x^2}{4!}+\frac{x^3}{5!}+...+\frac{x^n}{(n+2)!}+...\infty$
Hence fourth term is $\frac{x^3}{5!}$.