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LH
0
NT
5 tháng 9 2016
Ta biến đổi vế phải :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{49}-\frac{1}{50}\\ \)\(\\ =\left(1+\frac{1}{3}+\frac{1}{5}+........+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+.....+\frac{1}{50}\right)\\ =\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+.....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{25}\right)\\ =\frac{1}{26}+\frac{1}{27}+.....+\frac{1}{50}\)
Vậy \(\frac{1}{26}+\frac{1}{27}+.....+\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{49}-\frac{1}{50}\)
IM
5 tháng 9 2016
Ta có
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{50}\)
=> \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{50}\) ( đpcm )
31 tháng 3 2018
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{60}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-...-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\left(đpcm\right)\)
KH
15 tháng 5 2016
\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
\(< \frac{1}{26}+\frac{1}{26}+\frac{1}{26}+...+\frac{1}{26}+\frac{1}{26}\)
\(=\frac{25}{26}< 1\)(sai với đề bài)
22 tháng 7 2016
Ta có:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\left(đpcm\right)\)
T
24 tháng 4 2019
Ta có: \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}\) (đpcm)
*đpcm = điều phải chứng minh
14 tháng 5 2018
ta có :
\(VP=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-...-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}=VT\)
14 tháng 5 2018
Ta có: 1 - 1/2 + 1/3 - 1/4 + ... + 1/49 - 1/50
= 1 + 1/2 + 1/3 + 1/4 + ... + 1/49 + 1/50 - 2×( 1/2 + 1/4 + ... + 1/50)
= 1 + 1/2 + 1/3 + 1/4 + ... + 1/50 - (1 + 1/2 + 1/3 + ... + 1/25)
= 1/26 + 1/27 + 1/28 + ... + 1/49 + 1/50