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3 tháng 8 2018

Đc lém Min đúng lúc tui đang định đăng câu ó

3 tháng 8 2018

\(Ta\)  \(có\)  \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{256}\)

                   \(Vì\)  \(1>\frac{1}{256},\frac{1}{2}>\frac{1}{256},....,\frac{1}{255}>\frac{1}{256},\frac{1}{256}=\frac{1}{256}\)

                 \(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{256}>\frac{1}{256}+\frac{1}{256}+...+\frac{1}{256}\)

                  \(=\frac{1}{256}.256=1\)\(< 5\)

9 tháng 5 2017

Ta có :\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

=\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}=\)\(\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)\)\(+...+\left(1-\frac{1}{100}\right)\)

=\(\left(1+1+1+....+1\right)\)\(-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=             \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=  \(100-1-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=\(100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)= vế trên (đpcm)

9 tháng 5 2017

\(S=100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1+1+...+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(S=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(\RightarrowĐPCM\)

31 tháng 3 2020

\(\frac{1}{5}< \frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{1}{3}\)

+) Chứng minh: \(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{1}{3}\)

Có: \(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)

\(< \frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{3}-\frac{1}{100}< \frac{1}{3}\)

+) Chứng minh \(\frac{1}{5}< \frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)

\(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)

\(>\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{100.101}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{101}\)

\(=\frac{1}{4}-\frac{1}{101}=\frac{1}{5}+\frac{1}{20}-\frac{1}{101}>\frac{1}{5}\)

\(\frac{1}{5}< \frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{1}{3}\)

Trước hết ta phải chứng minh \(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{1}{3}\)

Ta có \(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)\(< \frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{3}-\frac{1}{100}< \frac{1}{3}\)

Sau đó chứng minh \(\frac{1}{5}< \frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)

\(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)\(>\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{100.101}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{101}\)

\(=\frac{1}{4}-\frac{1}{101}=\frac{1}{5}+\frac{1}{20}-\frac{1}{101}>\frac{1}{5}\)

Vậy .................

2 tháng 5 2016

Ta có: 1/2^2 < 1/1.2

          1/3^2 < 1/2.3 

        .........................

.......................................

          1/100^2 < 1/99.100

Ta có: 1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 1/1.2 + 1/2.3 + 1/3.4 + ...... + 1/99.100

          1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 1 - 1/100

          1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 99/100 < 3/4

         1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 3/4

2 tháng 5 2016

Ta có: 1/2^2 < 1/1.2

          1/3^2 < 1/2.3 

        .........................

.......................................

          1/100^2 < 1/99.100

Ta có: 1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 1/1.2 + 1/2.3 + 1/3.4 + ...... + 1/99.100

          1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 1 - 1/100

          1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 99/100 < 3/4

         1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 3/4

23 tháng 6 2020

Ta có:

\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

                                                                        \(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

                                                                        \(=\frac{1}{2}-\frac{1}{100}\)

                                                                       \(=\frac{49}{100}\)

Mà \(\frac{49}{100}< \frac{1}{2}\)

Vậy \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)

23 tháng 6 2020

Ta có:\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)(1)

Xét\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}\)

\(=\frac{50}{100}-\frac{1}{100}\)

\(=\frac{49}{100}\)(2)

\(\frac{49}{100}< \frac{50}{100}=\frac{1}{2}\)(3)

Từ (1), (2), (3)\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\left(đpcm\right)\)

Vậy...

Linz

22 tháng 2 2018

Ta có : 

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2010.2011}\)\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}=1-\frac{1}{2011}=\frac{2010}{2011}>\frac{2010}{2680}=\frac{3}{4}\)

Hình như có gì đó sai sai :')

22 tháng 2 2018

A+1/4=1/2+1/32+......+1/20112

A+1/4<1/2+1/2*3 +1/3*4 +....1/2010*2011

A+1/4<1-1/2011<1=3/4+1/4

A<1/4 (ĐPCM)

13 tháng 4 2017

Ta có: H=(1/2+1/3+1/4)+(1/5+...+1/8)+(1/9+1/16)+(1/17+...+1/63)

=> H=13/12 + (1/5+...+1/8)+(1/9+...+1/16)+(1/17+...+1/63)

=> H> 1 + 4x(1/8) + 8x (1/16) + (1/17+...+1/63)

=> H> 1+ 1/2 + 1/2 + (1/17+...+1/63)

=> H> 1+1+(1/17+...+1/63)

=> H>1+1

=> H>2