Question

Chứng minh :​ \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}< 1\).

Answer 1

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}=\frac{9}{10}< 1\)

\(\Rightarrowđpcm\)

Answer 2

Ta co: \(\frac{1}{2^2}< \frac{1}{1.2}\)

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

             ..................

         \(\frac{1}{10^2}< \frac{1}{9.10}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(\Rightarrow A< 1-\frac{1}{10}< 1\)

\(\Rightarrow A< 1\left(đpcm\right)\)

Answer 3

Ta có: \(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)

          \(\frac{1}{3^3}=\frac{1}{3.3}< \frac{1}{2.3}\)

                ........

      \(\frac{1}{10^2}=\frac{1}{10.10}< \frac{1}{9.10}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}< 1-\frac{1}{10}< 1\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}< 1\)

Vậy...

Answer 4

Vì :

\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)

\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)

\(................\)

\(\frac{1}{10^2}< \frac{1}{9\cdot10}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}< 1-\frac{1}{10}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}< \frac{9}{10}\)

mà \(\frac{9}{10}< 1\) \(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}< 1\).

Answer 5

Ta có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)=\(1-\frac{1}{10}=\frac{9}{10}< 1\)

Mà\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}=\)\(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{10.10}\)\(< \)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)< 1