A) \[lo{{g}_{e}}2\]
B) \[lo{{g}_{e}}2-1\]
C) \[2.lo{{g}_{e}}2\]
D) \[2.lo{{g}_{e}}2-1\]
Correct Answer: D
Solution :
\[\because \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}+.....to\,\infty .\] \[=\frac{2-1}{1.2}-\frac{3-2}{2.3}+\frac{4-3}{3.4}+.......up\,to\,\infty .\] \[=1-\frac{1}{2}-\frac{1}{2}+\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......up\,to\,\infty \] \[=\frac{2}{2}-\frac{2}{2}+\frac{2}{3}-\frac{1}{4}+\frac{2}{5}+.......up\,to\,\infty \] \[=1-2\left( \frac{1}{2}-\frac{1}{3}+-\frac{1}{4}++.....up\,to\,\infty \right)\] \[=1-21o{{g}_{e}}2\]You need to login to perform this action.
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