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1.3.5.....197.199 = \(\frac{\left(1.3.5.....197.199\right)\left(2.4.6.....198.200\right)}{2.4.6......198.200}\)= \(\frac{1.2.3......199.200}{2^{100}.\left(1.2.3.....100\right)}=\frac{101.102.103......200}{2^{100}}=\frac{101}{2}.\frac{102}{2}.\frac{103}{2}.....\frac{200}{2}\)
Ta có :
\(\frac{1}{101}>\frac{1}{200}\)
\(\frac{1}{102}>\frac{1}{200}\)
\(\frac{1}{103}>\frac{1}{200}\)
\(..........\)
\(\frac{1}{200}=\frac{1}{200}\)
Cộng vế với vế ta được :
\(\frac{1}{101}+\frac{1}{102}+....+\frac{1}{200}>\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\) (có 100 số \(\frac{1}{200}\) )\(=\frac{100}{200}=\frac{1}{2}\)
\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+......+\frac{1}{200}>\frac{1}{2}\) (đpcm)
Ta có:
1/101>1/200
1/102>1/200
...
1/199>1/200
=>1/101+1/102+...+1/103>1/200+1/200+...+1/200(100 số 1/200)
=1/200.100=1/2
Vậy 1/101+1/102+1/103+...+1/200>1/2
Ta có :
\(1.3.5.7.....199\)
\(=\frac{1.2.3.4.5.6.7.....198.199.200}{2.4.6.....198.200}\)
\(=\frac{\left(1.2.3.....99.100\right)\left(101.102.....200\right)}{\left(1.2.3.....99.100\right)\left(2.2.2.....2.2\right)}\)
\(=\frac{101.102.....200}{2.2.....2}\)
\(=\frac{101}{2}.\frac{102}{2}.....\frac{200}{2}\left(đpcm\right)\)
1/2=1/200+1/200+1/200+.....+1/200 (có 100 số )
1/101+1/102+....+1/200(có 100 số )
Vì 1/101>1/200
1/102>1/100
......
1/199>1/200
1/200=1/200
=>1/101+1/102+.....+1/200>1/200+1/200+...+1/200 có 100 số
=>1/101+1/102+.....+1/200>1/2
Ta thấy \(\frac{1}{101}>\frac{1}{200};\frac{1}{102}>\frac{1}{200};\frac{1}{103}>\frac{1}{200};....;\frac{1}{200}=\frac{1}{200}\)
Mà dãy \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+....+\frac{1}{200}\)có 100 phân số nên :
\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}>\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\)( có 100 phân số \(\frac{1}{200}\))
\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}>\frac{1}{200}.100=\frac{1.}{2}\left(đpcm\right)\)
Ta có:
\(\frac{1}{101}\)>\(\frac{1}{200}\)
\(\frac{1}{102}\)>\(\frac{1}{200}\)
\(\frac{1}{103}\)>\(\frac{1}{200}\)
...
\(\frac{1}{200}\)=\(\frac{1}{200}\)
\(\frac{1}{101}\)+\(\frac{1}{102}\)+\(\frac{1}{103}\)+...+\(\frac{1}{200}\)>\(\frac{1}{200}\)+\(\frac{1}{200}\)+..+\(\frac{1}{200}\)(100 số hạng)=\(\frac{1}{2}\)
\(\Rightarrow\)\(\frac{1}{101}\)+\(\frac{1}{102}\)+\(\frac{1}{103}\)+...+\(\frac{1}{200}\)>\(\frac{1}{2}\)
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Đặt \(S=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{199\cdot200}\)
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{199}-\frac{1}{200}\)
\(S=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(S=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Ta có đpcm
#)Giải :
Ta có : \(\frac{101}{2}.\frac{102}{2}.\frac{103}{2}.....\frac{200}{2}=\frac{101.102.103.....200}{2^{100}}=\frac{\left(101.102.103.....200\right)\left(1.2.3.....100\right)}{2^{100}\left(1.2.3.....100\right)}\)
\(=\frac{1.2.3.....200}{\left(2.1\right)\left(2.2\right)\left(2.3\right)...\left(2.100\right)}=\frac{\left(1.3.5.....99\right)\left(2.4.6.....100\right)}{2.4.6.....200}=1.3.5.....99\left(đpcm\right)\)
Ta có : 1.3.5.7.....199 = \(\frac{\left(1.3.5.7.....199\right).\left(2.4.6.8.....200\right)}{2.4.6.8.....200}=\frac{1.2.3.4.5.....199.200}{\left(1.2\right).\left(2.2\right).\left(3.2\right).....\left(100.2\right)}=\frac{1.2.3.4.5.....199.200}{2^{100}.1.2.3.....100}=\frac{101.102.103.....200}{2^{100}}\)\(=\frac{101}{2}.\frac{102}{2}\frac{103}{2}.....\frac{200}{2}\)\( \left(ĐPCM\right)\)