1/4x4+1/6x6+1/8x8+1/10x10+....+1/2nx2n < 1/4
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23 tháng 5 2016
1 + 2 x 2 + 3 x 3 + 4 x 4 + 5 x 5 + 6 x 6 + 7 x 7 + 8 x 8 + 9 x 9 + 10 x 10
= 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100
= 385
23 tháng 5 2016
1 + 2x2 + 3x3 + 4x4 + 5x5 + 6x6 + 7x7 + 8x8 + 9x9 + 10x10
= 1+4+9+16+25+36+49+64+81+100
=(81+9)+(64+16)+(49+1)+)36+4)+25+100
=90+80+50+40+25 +100
=385
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).
LN
13
26 tháng 8 2020
1 x 1 = 1
2 x 2 = 4
3 x 3 = 9
4 x 4 = 16
5 x 5 = 25
6 x 6 = 36
7 x 7 = 49
8 x 8 = 64
9 x 9 = 81
10 x 10 = 100
26 tháng 8 2020
\(1x1=1\)
\(2x2=4\)
\(3x3=9\)
\(4x4=16\)
\(5x5=25\)
\(6x6=36\)
\(7x7=49\)
\(8x8=64\)
\(9x9=81\)
\(10x10=100\)
LT
1
8 tháng 7 2015
đặt \(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+\frac{1}{7.7}+\frac{1}{8.8}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
\(A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=1-\frac{1}{8}<1\)
vậy \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+\frac{1}{7.7}+\frac{1}{8.8}<1\)
DH
0
PD
2
20 tháng 3 2015
ta đặt vế trái là A ta có:
A=1/2.2 .(1+1/2.2+1/3.3+1/4.4+...+1/50.50)
A< 1/2.2.(1+1/1.2+1/2.3+1/3.4+1/4.5+...+1/49.50)
A< 1/2.2.(1+1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+....+1/49-1/50)
A< 1/2.2.(1+1-150)
A< 1/2.2.99/50
A< 1/4.99/50
A< 99/200<100/200=1/2
=>A<1/2
DP
7 tháng 7 2017
\(\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+....+\frac{1}{10\cdot10}\)
Ta có :
\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)
.....................................
\(\frac{1}{10\cdot10}< \frac{1}{9\cdot10}\)
Ta có :
\(\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
\(\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< \frac{1}{1}-\frac{1}{10}\)
\(\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< \frac{9}{10}\)
\(\Rightarrow\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< \frac{9}{10}< 1\)
11 tháng 6 2018
Đặt \(B=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{10.10}\)
\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow B< 1-\frac{1}{10}< 1\)
\(\Rightarrow B< 1\left(đpcm\right)\)