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

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)

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

Ta có : \(\frac{1}{3^2}< \frac{1}{2.3}\)

            \(\frac{1}{4^2}< \frac{1}{3.4}\)

            \(\frac{1}{5^2}< \frac{1}{4.5}\)

             ...

            \(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

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

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

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

Ta có:

\(\frac{1}{3^2}

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

Ta có : Đặt A = \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

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

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

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

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

Vậy A < 1/2 

theo bài ra ta có:

1/3² <1/(2.3); 1/4²<1/(3.4);1/5²<1/(4.5);...;1/100²<1/(99.100)

=> 1/3^2+1/4^2+1/5^2+...+1/100^2  < 1/(2.3) + 1/3.4) +1/(4.5) +...+ 1/(99.100)   (1)

mà 1/(2.3)+1/(3.4) +1/(4.5) +...+ 1/(`99.100) = 1- 1/100= 99/100

ta có 99/100<1/2   (2)

từ (1) và (2)

=> điều phải CM

Đặt \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+....+\frac{1}{100^2}\)

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

\(=>A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}\)

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

\(=>A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)

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

Bạn xem lời giải của mình nhé:

Giải:

Gọi \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)

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

\(\Rightarrow A< \frac{1}{3}-\frac{1}{100}\\ \frac{1}{3}< \frac{1}{2}\Rightarrow\frac{1}{3}-\frac{1}{100}< \frac{1}{2}\\ \Rightarrow A< \frac{1}{2}\)

Chúc bạn học tốt!

Nguyễn Thế Bảo sai rồi,1/3^2<1/2.3 chứ sao là 1/3.4 đc?

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)

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

cunasai

Ta có: \(\frac{1}{5^2}< \frac{1}{4.5};\frac{1}{6^2}< \frac{1}{5.6};...;\frac{1}{100^2}< \frac{1}{99.100}\)

Cộng vế với vế ta được: \(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}< \frac{6}{24}=\frac{1}{4}\)(1)

Tương tự: \(\frac{1}{5^2}>\frac{1}{5.6};\frac{1}{6^2}>\frac{1}{6.7};...;\frac{1}{100^2}>\frac{1}{100.101}\)

Cộng vế với vế ta được \(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{96}{576}=\frac{1}{6}\)(2)

Từ (1) và (2) =>đpcm