Question

The sum of the series $\log_{4}2 - \log_{8}2 + \log_{16}2....$ is.

($\log_{4}2 - \log_{8}2 + \log_{16}2....$ dk ;ksx gSA)

• A. $e^{2}$
• B. $\log_{e}2$
• C. $\log_{e}3 - 2$
• D. $1 - \log_{e}2$

Answer 1

Answer: (D)

$1 - \log_{e}2$

Answer 2

Since $\frac{1}{2} + \frac{1}{4} + \frac{1}{8(2)!} + \frac{1}{16(3)!} + \frac{1}{32(4)!} + ......\infty =$and $e$.

$\sqrt{e}$

Also $\frac{\sqrt{e}}{2}$

Putting $\frac{2\frac{1}{2}}{1!} + \frac{3\frac{1}{2}}{2!} + \frac{4\frac{1}{2}}{3!} + \frac{5\frac{1}{2}}{4!} + ......\infty$.