JEE Main
- Exam
- JEE Main
- Subject
- Mathematics
- Chapter
- Basics of Mathematics
JEE Main · Mathematics · Basics of Mathematics
_4 2- _8 2+ _ 16 2- is equal to
is equal to
Options
- A
- B
- C
- D
Correct answer
D.
Step-by-step solution
4}-\frac{1}{\log _2 8}+\frac{1}{\log _2 16}-\ldots\left\{\because \log _b a=\frac{1}{\log _a b}\right\}=\frac{1}{\log _2(2)^2}-\frac{1}{\log _2(2)^3}+\frac{1}{\log _2(2)^4}-\ldots=\frac{1}{2 \log _2 2}-\frac{1}{3 \log _2 2}+\frac{1}{4 \log _2 2}-\ldots\left(\because \log _2 2=1\right)=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\ldots\right)\because \log _e(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots (Put )