Question

[nguyễn thị mai linh]

CHỨNG MINH RẰNG

1/3 ^ 2+1/4^2 + ............+1/100 ^2 nhỏ hơn 1/2

Answer 1

[Hiiiii~]

Có:

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

Mà: \(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

...

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

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

Mà \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{2}-0-0-...-0-\dfrac{1}{100}\)

\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)

Hay \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{49}{100}\)

Mà \(\dfrac{49}{100}>\dfrac{1}{2}\)

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

Chúc bạn học tốt!

Answer 2

[Nguyễn Thế Mãnh]

Ta có:

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

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

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

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