Sequence And Series Question 31
Question: $ \frac{1}{1,.,2,.,3}+\frac{1}{3,.,4.,5}+\frac{1}{5,.,6.,7}+…..\infty = $
Options:
A) $ {\log_{e}}\sqrt{2} $
B) $ {\log_{e}}2-\frac{1}{2} $
C) $ {\log_{e}}2 $
D) $ {\log_{e}}4 $
Show Answer
Answer:
Correct Answer: B
Solution:
We know $ {\log_{e}}2=( 1-\frac{1}{2} )+( \frac{1}{3}-\frac{1}{4} )+( \frac{1}{5}-\frac{1}{6} )+…….. $ ?..(i) $ =\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+ $ …….. ?..(ii) Again, $ {\log_{e}}2=1-( \frac{1}{2}-\frac{1}{3} )-( \frac{1}{4}-\frac{1}{5} )-………. $ $ =1-\frac{1}{2.3}-\frac{1}{4.5}-…….. $ ?..(iii) Add (ii) and (iii), $ 2.{\log_{e}}2=1+( \frac{1}{1.2}-\frac{1}{2.3} )+( \frac{1}{3.4}-\frac{1}{4.5} )+…… $ $ =(\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}{ 2{\log_{e}}2-1 }={\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}[ \frac{1}{2n-1}-\frac{1}{2n} ]-\frac{1}{2}[ \frac{1}{2n}-\frac{1}{2n+1} ] $ Putting $ n=1,2,3,…….. $ $ T_1=\frac{1}{2}[ \frac{1}{1}-\frac{1}{2} ]-\frac{1}{2}[ \frac{1}{2}-\frac{1}{3} ],\ T_2=\frac{1}{2}[ \frac{1}{3}-\frac{1}{4} ]-\frac{1}{2}[ \frac{1}{4}-\frac{1}{5} ] $ , $ T_3=\frac{1}{2}[ \frac{1}{5}-\frac{1}{6} ]-\frac{1}{2}[ \frac{1}{6}-\frac{1}{7} ] $ ……………………………………… ……………………………………… $ T_{n}=\frac{1}{2}[ \frac{1}{2n-1}-\frac{1}{2n} ]-\frac{1}{2}[ \frac{1}{2n}-\frac{1}{2n+1} ] $ Adding all terms, we get $ S_{n}=T_1+T_2+T_3+………+T_{n}+…….. $ $ =\frac{1}{2}[ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+…… ] $ $ -\frac{1}{2}[ \frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}+…… ] $ $ =\frac{1}{2}{\log_{e}}(1+1)+\frac{1}{2}[ -1+{ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+……. } ] $ $ =\frac{1}{2}{\log_{e}}2-\frac{1}{2}+\frac{1}{2}{\log_{e}}(1+1)={\log_{e}}2-\frac{1}{2} $ .
BETA
