ISI2016-DCG-23

Question

The value of $\log_{2}e-\log_{4}e+\log_{8}e-\log_{16}e+\log_{32}e-\cdots\:\:$ is

• A. $-1$
• B. $0$
• C. $1$
• D. None of these

Answer 1

Answer: $\mathbf C$

Explanation:

We know that:

$\color {blue} {\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3}}-\cdots\tag{1}$

Now, 

Let $\mathrm {V=\log_2e - \log_4e+\log_8e-\log_{16}e+\log_{32}e }\cdots$

$\mathrm {V=\log_2e-\log_{2^2}e+\log_{2^3}e-\log_{2^4}e}\cdots$

$\mathrm {V=\log_2e-\frac{1}{2}\log_{2}e+\frac{1}{3}\log_{2}e-\frac{1}{4}\log_{2}e} \cdots$

On taking out $\color {black} {\mathbf {\log_2e}}$ common, we get:

$\mathrm {V = \log_2e\left ( 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...\right ) } = \log_2e\times \log_e(1+1)\;\;\text{[From (1), here x = 1]} = \log_2e\times \log_e2\tag{2}$

We know that:

$\mathrm {\log_ab\times\log_ba = 1}\tag{3}$

From $(2)$ and $(3)$, we get:

$\mathrm V = 1$

$\therefore \mathbf C$ is the correct option.