Question

The value of $lo{g}_{4}2-lo{g}_{8}2+lo{g}_{16}2+...\infty$ is

• A. $e^2$
• B. ln 2+1
• C. ln 2-1
• D. 1-ln 2

Answer 1

The correct option is B ln 2+1

$lo{g}_{4}2-lo{g}_{8}2+lo{g}_{16}2+...\infty$
$\frac{lo{g}_{10}2}{lo{g}_{10}4}-\frac{lo{g}_{10}2}{lo{g}_{10}8}+\frac{lo{g}_{10}2}{lo{g}_{10}16}+...$
$\frac{lo{g}_{10}2}{lo{g}_{10}{2}^2}-\frac{lo{g}_{10}2}{lo{g}_{10}{2}^3}+\frac{lo{g}_{10}2}{lo{g}_{10}{2}^4}+...$
$\frac{lo{g}_{10}2}{2lo{g}_{10}2}-\frac{lo{g}_{10}2}{3lo{g}_{10}2}+\frac{lo{g}_{10}2}{4lo{g}_{10}2}+...$
$\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...$
$+1-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...\infty$
$-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+...+1$
$ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$
$ln\left(2\right)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}$
$-ln\left(2\right)=-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}$.
$-ln2+1$