Question

1/4^2 + 1/5^2 + 1/6^2 + ... + 1/64^2 < 5/16

Answer 1

Ta có: \(\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{64^2}< \frac{1}{4^2}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{63.64}\)

\(\frac{1}{4^2}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{63.64}=\frac{1}{4^2}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{63}-\frac{1}{64}\)

\(=\frac{1}{4^2}+\frac{1}{4}-\frac{1}{64}\)

VÌ: \(\frac{1}{4^2}+\frac{1}{4}-\frac{1}{64}< \frac{1}{4^2}+\frac{1}{4}=\frac{5}{16}\)

Nên: \(\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{64^2}< \frac{5}{16}\left(dpcm\right)\)