CM rằng
\(\frac{1}{6}< \frac{1}{5x5}+\frac{1}{6x6}+\frac{1}{7x7}+............+\frac{1}{100x100}< \frac{1}{4}\)
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Những câu hỏi liên quan
10 tháng 5 2017
Ta thấy:
1/2*2<1/1*2)vì 2*2>1*2).
1/3*3<1/2*3(vì 3*3>2*3).
...
1/8*8<1/7*8(vì 8*8>7*8).
=>1/2*2+1/3*3+1/4*4+...+1/8*8<1/1*2+1/2*3+1/3*4+...+1/7*8.
=>B<1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8.
=>B<1-1/8.
=>B<7/8.
Mà 7/8<1.
=>B<1.
Vậy B<1(đpcm).
PD
0
TM
2
7 tháng 6 2017
Ta thấy \(\frac{2}{3\times3},\frac{2}{5\times5},\frac{2}{7\times7},\frac{2}{9\times9}>0\)
\(\frac{2}{3\times3}=\frac{2}{9}=\frac{10}{45}>\frac{1}{45}\)
\(\Rightarrow M=\frac{2}{3\times3}+\frac{2}{5\times5}+\frac{2}{7\times7}+\frac{2}{9\times9}>\frac{1}{45}\)
HH
0
19 tháng 5 2018
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
AK
19 tháng 5 2018
Ta có :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\left(đpcm\right)\)
Chúc bạn học tốt !!!
10 tháng 3 2021
đặt A=1/5x5 +1/6x6 + 1/7x7 + .....+ 1/100x100
=>A>1/5x6 + 1/6x7 +1/7x8 + .... + 1/100x101
=>A>1/5 - 1/6 + 1/6 - 1/7 + +1/7 - 1/8 + ..... + 1/100 - 1/101
=>A> 1/5 - 1/101
=>A>96/505 > 96/576 = 1/6
=>A>1/6
=>A>B
10 tháng 3 2021
a>1/5x6+1/6x7+...+1/100x101
=1/5-1/6+1/6-1/7+...+1/100-1/101
=1/5-1/101
=101/505-5/101
=96/101
vì 96/101>1/6 nên a>1/6
Ta có:
1/5×5 < 1/4×5
1/6×6 < 1/5×6
1/7×7 < 1/6×7
.........
1/100×100 < 1/99×100
=> 1/5×5 + 1/6×6 + 1/7×7 +.....+ 1/100×100 < 1/4×5 + 1/5×6 + 1/6×7 +.....+ 1/99×100
= 1/4-1/5 + 1/5-1/6 + 1/6-1/7 +......+ 1/99-1/100
= 1/4-1/100 < 1/4
=> 1/5×5 + 1/6×6+1/7×7 +...+1/100×100<1/4 (1)
Lại có:
1/5×5 > 1/6×7
1/6×6 > 1/7×8
1/7×7 > 1/8×9
........
1/100×100 > 1/101×102
=> 1/5×5 + 1/6×6 + 1/7×7 +.....+ 1/100×100 > 1/5×6 + 1/6×7 + 1/7×8 +.....+1/100×101
= 1/5-1/6 + 1/6-1/7 + 1/7-1/8 +.....+ 1/100 - 1/101
= 1/5 - 1/101 > 1/5 - 1/30 = 1/6
=> 1/5×5 + 1/6×6 +1/7×7 +.....+ 1/100×100>1/6 (2)
Từ (1) và (2)
=> 1/6 < 1/5×5 +1/6×6+ 1/7×7 +...+1/100×100<1/4
Đặt \(A=\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}\)
Có \(\frac{1}{5.5}< \frac{1}{4.5};\frac{1}{6.6}< \frac{1}{5.6};...;\frac{1}{100.100}< \frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)(1)
Lại có :\(\frac{1}{5.5}>\frac{1}{5.6};\frac{1}{6.6}>\frac{1}{6.7};...;\frac{1}{100.100}>\frac{1}{100.101}\)
\(\Rightarrow A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\left(2\right)\)
Từ (1) và (2) \(\RightarrowĐCCM\)