Question

The value of $1-\log 2+\frac{{(\log 2)}^2}{2!}-\frac{{(\log )}^3}{3!}+...$ is

• A. $\log 3$
• B. $\log 2$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

Answer: (C)

In a given series there is a factorial in denominator, therefore it may be in the form of exponential series.
$\therefore$ $1-\log 2+\frac{{(\log 2)}^2}{2!}-\frac{{(\log 2)}^3}{3!}+...$
$=e^{-\log 2}$
$=e^{\log 2^{-1}}$
$=2^{-1}=\frac{1}{2}$