CMR:\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{100^2}< 0,74\)
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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 ~
1 tháng 4 2016
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+....................+\frac{1}{100}\right)\)
\(=100\cdot1-1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-..........................-\frac{1}{100}\)
\(=1-1+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+.......................+\left(1-\frac{1}{100}\right)\)
\(=0+\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+..................+\frac{99}{100}\left(ĐPCM\right)\)
1 tháng 4 2016
áp dụng quy tắc dấu ngoặc ta có: 100 - ( 1+1/2+1/3+...+1/100) = 100 - 1 - 1/2 - 1/3 - ...-1/100
=( 1-1/2)+(1-1/3)+(1-1/4)+...+(1-1/100) / có 100 số hạng
=1/2+2/3+3/4+...+99/100
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
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 :
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{16}\right)+...+\left(\frac{1}{81}-\frac{1}{100}\right)=1-\frac{1}{100}=\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
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)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
=\(\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
=\(\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)=\(1+1-\frac{1}{99}-\frac{1}{100}\)
=\(2-\frac{1}{99}-\frac{1}{100}< 2\)
\(\Rightarrow\)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
Ta có \(\frac{1}{3^2}< \frac{1}{2.3}\)
.....
\(\frac{1}{100^2}< \frac{1}{99.100}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{2^2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{100}=0,74\)
Tại sao sai z?