Question

Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\\ CMRA< \frac{3}{4}\)

Answer 1

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}=\frac{1}{4}+\left(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)

Ta có: \(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

\(\frac{1}{4^2}< \frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)

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

\(\frac{1}{100^2}< \frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}=\frac{3}{4}-\frac{1}{100}< \frac{3}{4}\)

Vậy A < 3/4

Answer 2

Thanks bạn nhiều