Question

 

A=1/2^2+1/3^2+1/4^2+.....+1/2018^2 với B=75%

So sánh A và B

Answer 1

\(\frac{1}{3^2}< \frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3},\frac{1}{4^2}< \frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4},...,\frac{1}{2018^2}< \frac{1}{2017\cdot2018}=\frac{1}{2017}-\frac{1}{2018}\)

\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}=\frac{1}{2}-\frac{1}{2018}< \frac{1}{2}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}< \frac{1}{2^2}+\frac{1}{2}=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}=75\%\)