A) \[{{\log }_{e}}\sqrt{2}\]
B) \[{{\log }_{e}}2-\frac{1}{2}\]
C) \[{{\log }_{e}}2\]
D) \[{{\log }_{e}}4\]
Correct Answer: B
Solution :
We know \[{{\log }_{e}}2=\left( 1-\frac{1}{2} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)+\left( \frac{1}{5}-\frac{1}{6} \right)+........\] ?..(i) \[=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\]........ ?..(ii) Again, \[{{\log }_{e}}2=1-\left( \frac{1}{2}-\frac{1}{3} \right)-\left( \frac{1}{4}-\frac{1}{5} \right)-..........\] \[=1-\frac{1}{2.3}-\frac{1}{4.5}-........\] ?..(iii) Add (ii) and (iii), \[2.{{\log }_{e}}2=1+\left( \frac{1}{1.2}-\frac{1}{2.3} \right)+\left( \frac{1}{3.4}-\frac{1}{4.5} \right)+......\] \[=(\alpha +\beta )x-\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{2}{{x}^{2}}+\frac{{{\alpha }^{3}}+{{\beta }^{3}}}{3}{{x}^{3}}-......\] ?..(iv) \[\Rightarrow \frac{1}{1.2.3}+\frac{1}{3.4.5}+......=\frac{1}{2}\left\{ 2{{\log }_{e}}2-1 \right\}={{\log }_{e}}2-\frac{1}{2}\] . Aliter: \[\frac{1}{1.2.3}+\frac{1}{3.4.5}+\frac{1}{5.6.7}+.......\] \[{{T}_{n}}=\frac{1}{(2n-1)(2n)(2n+1)}=\frac{1}{2(2n-1)}-\frac{1}{2n}+\frac{1}{2(2n+1)}\] \[=\frac{1}{2}\left[ \frac{1}{2n-1}-\frac{1}{2n} \right]-\frac{1}{2}\left[ \frac{1}{2n}-\frac{1}{2n+1} \right]\] Putting \[n=1,2,3,........\] \[{{T}_{1}}=\frac{1}{2}\left[ \frac{1}{1}-\frac{1}{2} \right]-\frac{1}{2}\left[ \frac{1}{2}-\frac{1}{3} \right],\ {{T}_{2}}=\frac{1}{2}\left[ \frac{1}{3}-\frac{1}{4} \right]-\frac{1}{2}\left[ \frac{1}{4}-\frac{1}{5} \right]\], \[{{T}_{3}}=\frac{1}{2}\left[ \frac{1}{5}-\frac{1}{6} \right]-\frac{1}{2}\left[ \frac{1}{6}-\frac{1}{7} \right]\] ............................................. ............................................. \[{{T}_{n}}=\frac{1}{2}\left[ \frac{1}{2n-1}-\frac{1}{2n} \right]-\frac{1}{2}\left[ \frac{1}{2n}-\frac{1}{2n+1} \right]\] Adding all terms, we get \[{{S}_{n}}={{T}_{1}}+{{T}_{2}}+{{T}_{3}}+.........+{{T}_{n}}+........\] \[=\frac{1}{2}\left[ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...... \right]\] \[-\frac{1}{2}\left[ \frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}+...... \right]\] \[=\frac{1}{2}{{\log }_{e}}(1+1)+\frac{1}{2}\left[ -1+\left\{ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....... \right\} \right]\] \[=\frac{1}{2}{{\log }_{e}}2-\frac{1}{2}+\frac{1}{2}{{\log }_{e}}(1+1)={{\log }_{e}}2-\frac{1}{2}\].
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