So sánh tổng sau vs \(\frac{1}{45}\)
M = \(\frac{2}{3x3}+\frac{2}{5x5}+\frac{2}{7x7}+\frac{2}{9x9}\)
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PD
0
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).
LL
6
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
29 tháng 4 2018
ta có: \(S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2007}{2^{2007}}\)
\(\Rightarrow\frac{1}{2}S=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+...+\frac{2007}{2^{2008}}\)
\(\Rightarrow S-\frac{1}{2}S=\frac{1}{2}+\left(\frac{2}{2^2}-\frac{1}{2^2}\right)+\left(\frac{3}{2^3}-\frac{2}{2^3}\right)+\left(\frac{4}{2^4}-\frac{3}{2^4}\right)+...+\left(\frac{2007}{2^{2007}}-\frac{2006}{2^{2007}}\right)-\frac{2007}{2^{2008}}\)
\(\frac{1}{2}S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2007}}-\frac{2007}{2^{2008}}\)
Gọi \(Q=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2007}}\)
\(\Rightarrow\frac{1}{2}Q=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2008}}\)
\(\Rightarrow Q-\frac{1}{2}Q=\frac{1}{2}-\frac{1}{2^{2008}}\)
\(\Rightarrow\frac{1}{2}Q=\frac{1}{2}-\frac{1}{2^{2008}}\)
\(Q=\left(\frac{1}{2}-\frac{1}{2^{2008}}\right):\frac{1}{2}=1-\frac{1}{2^{2007}}\)
Thay Q vào S, ta có:
\(\frac{1}{2}S=1-\frac{1}{2^{2007}}-\frac{2007}{2^{2008}}\)
\(\Rightarrow S=\left(1-\frac{1}{2^{2007}}-\frac{2007}{2^{2008}}\right):\frac{1}{2}\)
\(S=2-\frac{1}{2^{2006}}-\frac{2007}{2^{2007}}< 2\)
\(\Rightarrow S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2007}{2^{2007}}< 2\)
NV
19 tháng 6 2018
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
19 tháng 6 2018
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) \(\RightarrowĐCCM\)
LV
11 tháng 11 2017
\(S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+.....+\frac{n}{2^n}+......+\frac{2017}{2^{2017}}\)
Với n > 2 thì \(\frac{n}{2^n}=\frac{n+1}{2^{n-1}}-\frac{n+2}{2^n}\)
\(\frac{n+1}{2^{n-1}}=\frac{n+1}{2^n:2}=\frac{n+1}{\frac{2^n}{2}}=\frac{2^{\left(n+1\right)}}{2^n}\)
\(\frac{n+1}{2^{n-1}}-\frac{n+2}{2^n}=\frac{2^{n+2}}{2^n}-\frac{n+2}{2^n}\)
\(=\frac{2^{n+2}-n-2}{2^n}\)
\(=\frac{n}{2^n}\)
\(\Leftrightarrow S=\frac{1}{2}+\left(\frac{2+1}{2^{2-1}}-\frac{2+2}{2^2}\right)+.....+\frac{2016+1}{2^{2015}}-\frac{2018}{2^{2016}}\)
\(=\frac{2017+1}{2^{2016}}-\frac{2019}{2^{2017}}\)
\(S=\frac{1}{2}+\frac{3}{2}-\frac{2019}{2017}\)
\(S=2-\frac{2019}{2017}\)
\(\Leftrightarrow S=2-\frac{2019}{2017}< 2\)
Hay \(S< 2\)
28 tháng 4 2018
\(M=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\)
\(\Rightarrow2M=2+1+\frac{1}{2}+...+\frac{1}{2^{2015}}\)
\(\Rightarrow2M-M=2-\frac{1}{2^{2016}}< 2\)
=>ĐPCM
28 tháng 4 2018
ta có
M = 1+1/2 +...+1/22016
2M=2+1+1/2+...+1/22014+1/22015
2M-M=(2+1+1/2+...+1/22015)-(1+1/2 +...+1/22016)
M=2-1/22016<2
Vậy M<2
Ta thấy \(\frac{2}{3\times3},\frac{2}{5\times5},\frac{2}{7\times7},\frac{2}{9\times9}>0\)
\(\frac{2}{3\times3}=\frac{2}{9}=\frac{10}{45}>\frac{1}{45}\)
\(\Rightarrow M=\frac{2}{3\times3}+\frac{2}{5\times5}+\frac{2}{7\times7}+\frac{2}{9\times9}>\frac{1}{45}\)
\(\frac{1}{45}\) > m (tổng phép tính)