Question

Cho C= \(\frac{7}{3^2.4^2}+\frac{9}{4^2.5^2}+\frac{11}{5^2.6^2}+.....\frac{29}{14^2.15^2}\)

CMR: C<\(\frac{1}{5}\)

Answer 1

\(C=...\)

\(=\frac{1}{3^2}+\frac{1}{4^2}-\frac{1}{4^2}+\frac{1}{5^2}-\frac{1}{5^2}+...-\frac{1}{14^2}+\frac{1}{14^2}-\frac{1}{15^2}\)

\(=\frac{1}{3^2}-\frac{1}{15^2}\)

\(=\frac{1}{9}-\frac{1}{225}\)

Do \(\frac{1}{9}-\frac{1}{225}\)<\(\frac{1}{5}\)

\(=>C< \frac{1}{5}\)( ĐPCM )

Answer 2

C = ...

=> C = \(\frac{1}{3^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{5^2}+...+\frac{1}{14^2}-\frac{1}{15^2}\)

C = \(\frac{1}{3^2}-\frac{1}{15^2}\)

Ta thấy : \(\frac{1}{3^2}< \frac{1}{5}\Leftrightarrow\frac{1}{3^2}-\frac{1}{15^2}< \frac{1}{5}\)

=> C < \(\frac{1}{5}\)