Question

\(CMR:\)\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}< 1\)

Answer 1

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\)(do 2²>1.2)

          \(\frac{1}{3^2}< \frac{1}{2.3}\)(do 3²>2.3)

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

           \(\frac{1}{8^2}< \frac{1}{7.8}\)(do 8²>7.8)

=> B < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\)

=> B < \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)
=> B < \(1-\frac{1}{8}< 1\)

=> B < 1

Answer 2

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

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

           ....................

           \(\frac{1}{8^2}< \frac{1}{7.8}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{7.8}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< 1-\frac{1}{8}=\frac{7}{8}< 1\left(đpcm\right)\)

Answer 3

Ta có:\(B=\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}=1-\frac{1}{8}< 1\)

=>\(B< 1\)

Vậy\(B< 1\)