Question

[Đặng Vũ Cường]

giúp mình với

Chứng minh nếu C = 1/5^2 + 1/6^2 + 1/7^2 +...+ 1/100^2 thì 1/6 < C < 1/4

Answer 1

[ST]

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

\(\frac{1}{6^2}< \frac{1}{5.6}\)

\(\frac{1}{7^2}< \frac{1}{6.7}\)

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

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

=> \(C< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{99.100}\)

=> \(C< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)

=> \(C< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)                                                             (1)

Lại có: \(\frac{1}{5^2}>\frac{1}{5.6}\)

\(\frac{1}{6^2}>\frac{1}{6.7}\)

\(\frac{1}{7^2}>\frac{1}{7.8}\)

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

\(\frac{1}{100^2}>\frac{1}{100.101}\)

=> \(C>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+....+\frac{1}{100.101}\)

=> \(C>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}\)

=> \(C>\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\)                                                                (2)

Từ (1) và (2) suy ra \(\frac{1}{6}< C< \frac{1}{4}\)(đpcm)