A) \[lo{{g}_{e}}2-1\]
B) \[1-lo{{g}_{e}}2\]
C) \[1+lo{{g}_{e}}2\]
D) e
Correct Answer: B
Solution :
[b] \[\because lo{{g}_{4}}^{2}-lo{{g}_{^{8}}}^{2}-lo{{g}_{16}}^{2}+.....up\,to\,\infty \] \[=\frac{1}{{{\log }_{1}}^{4}}-\frac{1}{{{\log }_{2}}^{8}}+\frac{1}{{{\log }_{2}}^{16}}+.....up\,to\,\infty \] \[=\frac{1}{4}-\frac{1}{3}+\frac{1}{4}+....up\,to\,\infty \] \[=1-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+.....up\,to\,\infty \] \[=1-\left( 1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4} \right)+......up\,to\,\infty \] \[=1-log\left( 1+1 \right)=1-log2.\] Hence, option [b] is correct.
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