A) \[\log \sqrt{12}\]
B) \[\log \sqrt{10}\]
C) \[{{e}^{12}}\]
D) None of these
Correct Answer: A
Solution :
We have the series \[1+\left( \frac{1}{2}+\frac{1}{3} \right)\frac{1}{4}+\left( \frac{1}{4}+\frac{1}{5} \right)\frac{1}{{{4}^{2}}}\] \[+\left( \frac{1}{6}+\frac{1}{7} \right)\frac{1}{{{4}^{3}}}+...\infty \] \[=\left( \frac{1}{2}.\frac{1}{4}+\frac{1}{4}.\frac{1}{{{4}^{2}}}+\frac{1}{6}.\frac{1}{{{4}^{3}}}+...\infty \right)\] \[+\left( 1+\frac{1}{3}.\frac{1}{4}+\frac{1}{5}.\frac{1}{{{4}^{2}}}+\frac{1}{7}.\frac{1}{{{4}^{3}}}+...\infty \right)\] \[=\frac{1}{2}\left[ \frac{1}{4}+\frac{1}{2}.\frac{1}{{{4}^{2}}}+\frac{1}{3}.\frac{1}{{{4}^{3}}}+...\infty \right]\] \[+2\left[ \frac{1}{2}+\frac{1}{3}{{\left( \frac{1}{2} \right)}^{3}}+\frac{1}{5}{{\left( \frac{1}{2} \right)}^{5}}+...\infty \right]\] Putting \[\frac{1}{4}=x\]and \[\frac{1}{2}=y\]in above series, we get the series \[\frac{1}{2}\left[ x+\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}+...\infty \right]\] \[+\,2\left[ y+\frac{{{y}^{3}}}{3}+\frac{{{y}^{5}}}{5}+...\infty \right]\] \[=-\frac{1}{2}\log (1-x)+\log \left( \frac{1+y}{1-y} \right)\] \[=-\frac{1}{2}\log \left( 1-\frac{1}{4} \right)+\log \left( \frac{1+\frac{1}{2}}{1-\frac{1}{2}} \right)\] \[=-\frac{1}{2}\log \frac{3}{4}+\log 3\] \[=\log \sqrt{\frac{4}{3}}+\log 3=\log \left( 3\sqrt{\frac{4}{3}} \right)\] \[=\log \sqrt{12}\]
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