Question

cho A=\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.......+\dfrac{1}{50^2}\) chứng minh A<2

giúp mình với đây là đề ôn thi hk2 chỉ là đề cương thui nha

Answer 1

\(\)Ta có: \(\dfrac{1}{1^2}=1;\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{50^2}< \dfrac{1}{49.50}\)

\(\Rightarrow A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

\(\Rightarrow A< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(\Rightarrow A< 1+\dfrac{99}{100}\)

\(\Rightarrow A< 1\dfrac{99}{100}< 1\dfrac{100}{100}=2\)

\(\Rightarrow A< 2\left(đpcm\right)\)