Question

[lê xân sơn]

CMR: A=3/2^2+8/3^2+...+2023^2 -1/2023^2 ko phải là 1 số tự nhiên

giúp mk vs chiều thi rồi!!!!!!

Answer 1

[Nguyễn Lê Phước Thịnh]

\(A=\dfrac{3}{2^2}+\dfrac{8}{3^2}+...+\dfrac{2023^2-1}{2023^2}\)

\(=\dfrac{2^2-1}{2^2}+\dfrac{3^2-1}{3^2}+...+\dfrac{2023^2-1}{2023^2}\)

\(=1+1+...+1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}\right)\)

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\dfrac{1}{2023^2}< \dfrac{1}{2022\cdot2023}=\dfrac{1}{2022}-\dfrac{1}{2023}\)

Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\)

=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< 1\)

=>\(0< \dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< 1\)

=>A không là số tự nhiên

Answer 2

[Nguyễn Bình]

A=3/2^2 + 8/3^2 + ... + 2023^2 - 1/2023^2

A =2^2-1/2^2  + 3^2-1/3^2 +...+ 2023^2-1/2023^2

A=1 - 1/2^2 + 1- 1/3^2 + ... + 1 - 1/2023^2

A=1+1+...+1 - (1/2^2 +1/3^2 + 1/4^2 +...+1/2023^2)

A=2022 - (1/2^2 + 1/3^2 + ... + 1/2023^2) <2022 (1)

Ta có 1/2^2 < 1/1.2

           1/3^2 <1/2.3

           .................

            1/2023^2 < 1/2022.2023

suy ra 

1/2^2 + 1/3^2 + ... +1/2023^2 <1/1.2 + 1/2.3 +...+1/2022.2023

Ta có 

1/1.2 + 1/2.3 + .... +1/2022.2023

=1/1 - 1/2 + 1/2 - 1/3 + ....+1/2022 - 1/2023

=1/1 - 1/2023

suy ra 1/2^2 + 1/3^2 + ... + 1/2023^2<1-1/2023

suy ra A =2022 - (1/2^2 + 1/3^2 + .... + 1/2023^2) > 2022-(1-2023)

suy ra 2022 - (1/2^2 + 1/3^2 +...+1/2023^2) >2021 + 1/2023 >2021(2)

tù 1,2 suy ra 

    2021<A<2022

 suy ra A ko là số tự nhiên 

Vậy A ko là số tự nhiên