Question

[Su Kem]

Chứng minh rằng

1/3²+1/4²+1/5²+1/6²...+1/100²<1/2

Answer 1

[Đức Minh]

Đặt \(B=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

\(\Rightarrow B< \dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{99\cdot100}\)

\(\Rightarrow B< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow B< \dfrac{1}{2}-\dfrac{1}{100}\left(1\right)\)

mà \(\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\left(2\right)\)

\(\left(1\right),\left(2\right)\rightarrow B< \dfrac{1}{2}\left(đpcm\right)\)

Answer 2

[Nguyễn Thế Mãnh]

Ta có:

\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}-\dfrac{1}{100}\)

Mà \(\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\)

=> \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\left(đpcm\right)\)