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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có 1+5/28=33/28
Đặt A=1/11+1/12+1/13+...+1/69+1/70
A=(1/11+1/12++1/13+...+1/20)+(1/21+1/22+1/23+...+1/30)+(1/31+1/32+1/33+...1/60)+...+1/70
Ta thấy :
1/11+1/12+1/13+...+1/20>1/20+1/20+1/20+...+1/20(có 10 số hạng 1/20)=1/20*10=1/2
1/21+1/22+1/23+...+1/30>1/30+1/30+1/30+...+1/30(10 số hạng 1/30)=1/30*10=1/3
1/30+1/31+1/32+...+1/60>1/60+1/60+...+1/60(30 số hạng 1/60)=1/60*30=1/2
1/61+1/62+1/63+...+1/70>1/70+1/70+1/70+...+1/70(10 số hạng 1/70)=1/70*10=1/7
=>1/11+1/12+1/13+...+1/69+1/70>1/2+1/3+1/2+1/7
=>A>31/21
Mà 31/21>33/28
=>A>33/28
=>A>1+5/28(DPCM)
Vậy A>1+5/28
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 11 + 112 + 113 + ... + 117 + 118
= (11 + 112) + (113 + 114) + ... + (117 + 118)
= 11(1 + 11) + 113(1 + 11) + ... + 117(1 + 11)
= 11.12 + 113.12 + .... + 117.12
= 12(11 + 113 + ... + 117) chia hết cho 12
b, 7 + 72 + 73 + 74
= (7 + 73) + (72 + 74)
= 7(1 + 72) + 72(1 + 72)
= 7.50 + 72.50
= 50(7 + 72) chia hết cho 50
c, 3 + 32 + 33 + 34 + 35 + 36
= (3 + 32 + 33) + (34 + 35 + 36)
= 3(1 + 3 + 32) + 34(1 + 3 + 32)
= 3.13 + 34.13
= 13(3 + 34) chia hết cho 13
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A = (\frac{1}{10} + ...+ \frac{1}{19} ) + (\frac{1}{20} + ...+ \frac{1}{29}) + (\frac{1}{30} +...+ \frac{1}{39} ) + (\frac{1}{40} + ...+\frac{1}{49} ) + (\frac{1}{50} +....+ \frac{1}{59}) + (\frac{1}{60} + ....+\frac{1}{69}) + \frac{1}{70}\)
Ta có : mỗi bên có 10 số hạng
\( (\frac{1}{10} + ..+ \frac{1}{19}) < (\frac{1}{10} + ...+ \frac{1}{10}) = \frac{1}{1}\)
\(\frac{1}{20}+..+ \frac{1}{29} < (\frac{1}{20}+..+\frac{1}{20}) = \frac{1}{2}\)
\((\frac{1}{30} +...+ \frac{1}{39} )< (\frac{1}{30} +...+ \frac{1}{30}) = \frac{1}{3}\)
\((\frac{1}{40} + ...+\frac{1}{49} )< (\frac{1}{40} + ...+\frac{1}{40}) = \frac{1}{4}\)
\((\frac{1}{50} +....+ \frac{1}{59})< (\frac{1}{50} +....+ \frac{1}{50}) = \frac{1}{5}\)
\((\frac{1}{60} + ....+\frac{1}{69}) + \frac{1}{70}< (\frac{1}{60} + ....+\frac{1}{60})+ \frac{1}{70} = \frac{1}{6} +\frac{1}{70}\)
\(\implies A < 1+\frac{1}{2} + ...+ \frac{1}{6} + \frac{1}{70}= \frac{13}{15} + \frac{1}{70} <1<\frac {51}{20} \)
\(\implies A<\frac{51}{20}\) \((đpcm)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
thực ra nó rất là dễ. giờ mình mới phát hiện ra chứ bữa trước mình làm cách dài lắm
Ta có :
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)\)
\(=\frac{25}{12}+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)>\frac{25}{12}\)( đpcm )
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{3}{1^2.2^2}=\frac{3}{1.4}=1-\frac{1}{4}\); \(\frac{5}{2^2.3^2}=\frac{5}{4.9}=\frac{1}{4}-\frac{1}{9}\); \(\frac{7}{3^2.4^2}=\frac{7}{9.16}=\frac{1}{9}-\frac{1}{16}\); ...; \(\frac{39}{19^2.20^2}=\frac{39}{361.400}=\frac{1}{361}-\frac{1}{400}\)
Gọi tổng đó là A => A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{361}-\frac{1}{400}\)
=> \(A=1-\frac{1}{400}=\frac{399}{400}< \frac{400}{400}=1\)
=> A < 1
Gọi \(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{69}+\dfrac{1}{70}\) là \(S\)
Ta nhận thấy:
\(\dfrac{1}{11},\dfrac{1}{12},\dfrac{1}{13},...,\dfrac{1}{19}\)đều lớn hơn \(\dfrac{1}{20}\)
\(\dfrac{1}{21},\dfrac{1}{22},\dfrac{1}{23},...,\dfrac{1}{29}\)đều lớn hơn \(\dfrac{1}{30}\) \(\dfrac{1}{31},\dfrac{1}{32},\dfrac{1}{33},...,\dfrac{1}{39}\)đều lớn hơn \(\dfrac{1}{40}\) \(\dfrac{1}{41},\dfrac{1}{42},\dfrac{1}{43},...,\dfrac{1}{49}\)đều lớn hơn \(\dfrac{1}{50}\) \(\dfrac{1}{51},\dfrac{1}{52},\dfrac{1}{53},...,\dfrac{1}{59}\)đều lớn hơn \(\dfrac{1}{60}\)\(\dfrac{1}{61},\dfrac{1}{62},\dfrac{1}{63},...,\dfrac{1}{69}\)đều lớn hơn \(\dfrac{1}{70}\)
\(\Rightarrow S< \dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}+\dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}+\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}+\dfrac{1}{70}+\dfrac{1}{70}+...+\dfrac{1}{70}\\ \Leftrightarrow S< \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}\\ =\dfrac{223}{140}\) \(1\dfrac{5}{29}=\dfrac{34}{29}\) \(\dfrac{223}{140}>\dfrac{210}{140}=\dfrac{3}{2}=\dfrac{87}{58}>\dfrac{34}{29}\) Vậy \(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{69}+\dfrac{1}{70}>1+\dfrac{5}{29}\left(đpcm\right)\)