Sequence And Series Question 34
Question: $ \frac{1}{2}+\frac{3}{2},.,\frac{1}{4}+\frac{5}{3}.\frac{1}{8}+\frac{7}{4}.\frac{1}{16}+…..\infty = $
Options:
A) $ 2-{\log_{e}}2 $
B) $ 2+{\log_{e}}2 $
C) $ {\log_{e}}4 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{1}{1}.\frac{1}{2}+\frac{3}{2}.\frac{1}{2^{2}}+\frac{5}{3}.\frac{1}{2^{3}}+\frac{7}{4}.\frac{1}{2^{4}}+……… $ $ =( 2-\frac{1}{1} )\frac{1}{2}+( 2-\frac{1}{2} )\frac{1}{2^{2}}+( 2-\frac{1}{3} )\frac{1}{2^{3}}+( 2-\frac{1}{4} )\frac{1}{2^{4}}+.. $ $ =2{ \frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+…….. }-{ \frac{\frac{1}{2}}{1}+\frac{\frac{1}{2^{2}}}{2}+\frac{\frac{1}{2^{3}}}{3}+……. } $ $ =\frac{1}{1-\frac{1}{2}}-{ -{\log_{e}}( 1-\frac{1}{2} ) }=2-{\log_{e}}2 $ .
BETA
