Question

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{2012^2}\)

Answer 1

Ta có: M=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2012^2}\)

\(\Rightarrow M< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\)

\(\Rightarrow M< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\)

\(\Rightarrow M< 1-\dfrac{1}{2012}\)

\(\Rightarrow M< 1\)

Vậy M<1

Answer 2

Chia buồn nhé, tui có câu trả lời rùi nè, muốn coi hông.