Question

Đề bài này đúng hay đề bài sai:

\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+.....+\(\frac{1}{2017^2}\)<\(\frac{3}{4}\)

Answer 1

Đề đúng!

Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2017^2}\)

Ta có: \(\frac{1}{2^2}=\frac{1}{4}\)

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

\(\frac{1}{4^2}< \frac{1}{3.4}\)

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

\(\frac{1}{2017^2}< \frac{1}{2016.2017}\)

\(\Rightarrow A< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)

\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)

\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{2017}=\frac{3}{4}-\frac{1}{2017}< \frac{3}{4}\)

Vậy A < 3/4

Answer 2

dề đúng