Question

lf $y=x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots .\infty$ then $y-\frac{1}{2}{y}^2+\frac{1}{3}{y}^3-\frac{1}{4}{y}^4+\dots .\infty =$

• A. $4x$
• B. $3x$
• C. $2x$
• D. $x$

Answer 1

Answer: (D)

$x$

Given, $y=x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots .\infty$
Add 1 to both sides,
$\Rightarrow 1+y=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots .\infty =e^x$
$\therefore x=\log (1+y)=y-\frac{1}{2}{y}^2+\frac{1}{3}{y}^3-............\infty$