Question

chứng minh rằng tổng S=1/1x3+1/3x5+1/5x7+.............+1/2015x2017<1/2

Answer 1

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)

\(S=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(S=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)

\(S=\frac{1}{2}.\frac{2016}{2017}\)

\(S=\frac{1008}{2017}< \frac{1}{2}\)

Answer 2

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)

\(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)

\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)

\(2S=1-\frac{1}{2017}< 1\)

=> 2S < 1 

=> S < \(\frac{1}{2}\)(đpcm)

Answer 3

S = \(\frac{1}{2}\). \(\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2015.2017}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2017}\right)\)= \(\frac{1}{2}.\frac{2016}{2017}\)= \(\frac{1008}{2017}< \frac{1008}{2016}=\frac{1}{2}\)
(=) \(S< \frac{1}{2}\)(đpcm)