Question

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

• A. $\frac{x}{3}$
• B. $\frac{x}{2}$
• C. $x$
• D. $2x$

Answer 1

Answer: (C)

$x$

Given $y=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots \infty =\log (1+x)$
$\Rightarrow 1+x=e^y=1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+.......$
Hence, $y+\frac{y^2}{2!}+\frac{y^3}{3!}+.......=x$