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Đặt A = 1-1/2+1/3-1/4 +...+1/1989-1/1990
A= (1+1/3+1/5 +...+1/1989)- ( 1/2 + 1/4 +....+1/1990 )
A=(1+1/3+1/5 +...+1/1989) - 2(1/2+1/4+1/6+.....+1/1990)
A= (1+1/3+1/5 +...+1/1989)- (1+1/2+1/3+1/4 +...+1/995)
A= 1/996+1/997 +.....+1/1990 =VP (đpcm)
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A=
xét vế trái
=(1+1/3+1/5+...+1/1989)-(1/2+1/4+...+1/1990)
=(1+1/2+1/3+1/4+...+1/1990)-2.(1/2+1/4+...+1/1990)
=(1+1/2+1/3+1/4+...+1/1990)-!1+1/2+1/3+1/4+...+1/995)
=1/996+1/997+.../1+1990
vậy 1-1/2+1/3-1/4+...-1/1990=1/996+1/997+...+1/1990
cmr 1-$\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.......-\frac{1}{1990}=\frac{1}{996}+\frac{1}{997}+\frac{1}{998}+.......+\frac{1}{1990}$
Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)(đpcm)
Ta có:
\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Mà \(\frac{49}{100}< \frac{1}{2}\)
Vậy \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
Ta có:\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)(1)
Xét\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{50}{100}-\frac{1}{100}\)
\(=\frac{49}{100}\)(2)
Mà\(\frac{49}{100}< \frac{50}{100}=\frac{1}{2}\)(3)
Từ (1), (2), (3)\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\left(đpcm\right)\)
Vậy...
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