Question

The value of $(x-1)-\frac{1}{2}(x-1{)}^2+\frac{1}{3}(x-1{)}^3-\frac{1}{4}(x-1{)}^4+\dots .$ is

• A. ${\log }_{e}x$
• B. $2{\log }_{e}x$
• C. $3{\log }_{e}x$
• D. $4{\log }_{e}x$

Answer 1

The correct option is A ${\log }_{e}x$

The expansion for $\log (1+x)$ is
$\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-....$
Substituting $x=x-1$ we get
$\log (1+x-1)={\log }_e\left(x\right)=(x-1)-\frac{(x-1{)}^2}{2}+\frac{(x-1{)}^3}{3}-\frac{(x-1{)}^4}{4}+\frac{(x-1{)}^5}{5}-....$