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1/1.2 + 1/2.3 + ...... + 1/49.50
= 1/1 - 1/2 + 1/2 - - .... - 1/50 = 1 - 1/50 = 49/50
Ta có: \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\)
\(-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
Ta có :
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
=> đpcm
Ủng hộ mk nha !!! ^_^
\(\text{Ta có :}\)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=\frac{1}{1}-\frac{1}{50}\)
\(A=\frac{49}{50}\)
Vì \(\frac{245}{420}< \frac{245}{294}< \frac{245}{250}\)
Vậy \(\frac{7}{12}< \frac{49}{50}< \frac{5}{6}\)
Ta có :
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-2.\frac{1}{2}-2.\frac{1}{4}-2.\frac{1}{6}-...-2.\frac{1}{50}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
Vậy ......
~Hok tốt~
CỨ TÍNH VẾ TRÁI R ĐC
1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50
(1+1/3+1/5+...+1/49)+(1+1/2+1/4+...+1/50)-(1/2+1/4+1/6+...+1/50)-(1/2+1/4+1/6+...+1/50)
(1+1/2+1/3+1/4+...+1/50)-2(1/2+1/4+1/6+..+1/50)
(1+1/2+1/4+1/6+...+1/50)-(1+1/2+1/3+..+1/25)
1/26+1/27+...+1/50
1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+...+1/50
=1/1-1/2+1/3-1/4+...+1/49-1/50
=(1/1+1/3+...+1/49)-(1/2+1/4+...+1/50)
=(1/1+1/2+1/3+...+1/49+1/50)-2(1/2+1/4+...+1/50)
=1/1+1/2+1/3+...+1/50-1-1/2-1/3-...-1/25
=1/26+1/27+...+1/50 (đpcm)
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