\(G=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) CMR G nhỏ hơn \(\frac{3}{4}\)
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PB
2
23 tháng 2 2020
Ta có:\(\frac{1}{2^2}=\frac{1}{4};\frac{1}{3^2}< \frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3};\frac{1}{3^2}< \frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4};.....;\frac{1}{100^2}< \frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}< \frac{3}{4}\left(đpcm\right)\)
23 tháng 2 2020
Gọi \(D=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}< \frac{3}{4}\)
Vì \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{100^2}< \frac{1}{99.100}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< \frac{3}{4}\)
\(\Rightarrow D< \frac{3}{4}\left(đpcm\right)\)
NT
27 tháng 8 2016
\(E=1-\frac{1}{2^2}-\frac{1}{3^2}-..........-\frac{1}{2004^2}\)
\(E=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+..........+\frac{1}{2014^2}\right)\)
Ta có : \(E< 1-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{2003.2004}\right)\\ \)
Đặt A= \(1-\left(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2003.2004}\right)\\ =>A=1-\left(1-\frac{1}{2004}\right)\\ =>A=1-\frac{2003}{2004}\\ =>A=\frac{1}{2004}\)
Chắc chắn bạn đã ghi nhầm dấu
14 tháng 1 2018
Có : (1+1/2+1/3+....+1/100)+(1/2+2/3+....+99/100)
= 1+(1/2+1/2)+(1/3+2/3)+.....+(1/100+99/100) ( có 99 cặp )
= 1+1+1+....+1 ( có 100 số 1 )
= 100
=> 100-(1+1/2+1/3+....+1/100)=1/2+2/3+3/4+....+99/100
Tk mk nha
6 tháng 10 2018
\(VP=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(VP=\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)
\(VP=\frac{2}{2}-\frac{1}{2}+\frac{3}{3}-\frac{1}{3}+\frac{4}{4}-\frac{1}{4}+...+\frac{100}{100}-\frac{1}{100}\)
\(VP=1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+...+1-\frac{1}{100}\)
\(VP=100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=VT\) ( đpcm )
Mk nghĩ \(VT=100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\) bn xem lại đề có nhầm ko
Chúc bạn học tốt ~
T
21 tháng 7 2019
#)Giải :
Bài 1 :
\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\Leftrightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(\Leftrightarrow3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(\Leftrightarrow2C=1-\frac{1}{3^{100}}\Leftrightarrow C=\frac{1-\frac{1}{3^{100}}}{2}< \frac{1}{2}\Rightarrow C< \frac{1}{2}\left(đpcm\right)\)
Bài 2 :
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{16}\right)+...+\left(\frac{1}{81}-\frac{1}{100}\right)=1-\frac{1}{100}=\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
14 tháng 8 2017
câu g)
\(G=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right).\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{120}{121}\)
\(=\frac{3.\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{2.2.3.3.4.4.5.5....11.11}\)
\(=\frac{12}{3}=4\)
VH
1
DY
13 tháng 3 2019
ta có : 1/2^2<1/2x3
1/3^2<1/3x4
...........
1/100^2<1/99x100
suy ra :1/2^2+1/3^2 +........+1/100^2<1/2x3+1/3x4+1/4x5+..........+1/99x100
Gọi A=1/2x3+1/3x4+............+1/99x100
A=3-2/2x3+4-3/3x4+..........+100-99/99x100
A=3/2x3-2/2x3+4/3x4-3/3x4+........+100/99x100-99/99x100
A=1/2-1/100
A=49/100
1/2^2+1/3^2+......+1/100^2<49/100
Ta có:3/4=75/10049/100
Mà 75/100>49/100
1/2^2+1/3^2+........+1/100^2<3/4
26 tháng 6 2019
a)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
=\(\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{100}{100!}-\frac{1}{100!}\)
=\(1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
=\(1-\frac{1}{100!}< 1\)
\(\Rightarrow\)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)
b)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
=\(\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
=\(\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)=\(1+1-\frac{1}{99}-\frac{1}{100}\)
=\(2-\frac{1}{99}-\frac{1}{100}< 2\)
\(\Rightarrow\)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)