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Đặt tổng là S
\(\Rightarrow S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\)
\(\Rightarrow S=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(\Rightarrow S=\left(1+\frac{1}{2}+....+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+....+\frac{1}{1008}\right)\)
\(\Rightarrow S=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\) (đpcm)
2.
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}\)
A = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)
=\(\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{2005}\right)\)\(-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
= \(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}+\frac{1}{2006}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
= \(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}+\frac{1}{2006}\)\(-\frac{1}{1}-\frac{1}{2}-...-\frac{1}{1003}\)
= \(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2005}+\frac{1}{2006}\)
(=) B - A = \(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}+\frac{1}{2016}\)- \(\frac{1}{1004}-\frac{1}{1005}-...-\frac{1}{2005}-\frac{1}{2006}\)
= \(\frac{1}{2007}+\frac{1}{2008}+...+\frac{1}{2016}-\) \(\frac{1}{1004}-\frac{1}{1005}-\frac{1}{1006}-\frac{1}{1007}\)
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
=> đpcm
Ủng hộ mk nha ^_-