A=1−2−3+4−5−6+7−8−9+....+2020−2021−2022D=1-2-3+4-5-6+7-8-9+....+2020-2021-2022

A =(1−2−3)+(4−5−6)+(7−8−9)+....+(2020−2021−2022)D=(1-2-3)+(4-5-6)+(7-8-9)+....+(2020-2021-2022)

A=(−4)+(−7)+(−10)+.....+(−2023)D=(-4)+(-7)+(-10)+.....+(-2023)

A=[(2023−4):3+1].[(−2023−4):2]D=[(2023-4):3+1].[(-2023-4):2]

A=674.(−1013,5)D=674.(-1013,5)

A=−683099

A=1−2−3+4−5−6+7−8−9+....+2020−2021−2022D=1-2-3+4-5-6+7-8-9+....+2020-2021-2022

A =(1−2−3)+(4−5−6)+(7−8−9)+....+(2020−2021−2022)D=(1-2-3)+(4-5-6)+(7-8-9)+....+(2020-2021-2022)

A=(−4)+(−7)+(−10)+.....+(−2023)D=(-4)+(-7)+(-10)+.....+(-2023)

A=[(2023−4):3+1].[(−2023−4):2]D=[(2023-4):3+1].[(-2023-4):2]

A=674.(−1013,5)D=674.(-1013,5)

A=−683099

Đặt

A= \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}=\frac{1}{\left(2.2\right)^2}+\frac{1}{\left(2.3\right)^2}+\frac{1}{\left(2.4\right)^2}+...+\frac{1}{\left(2n\right)^2}\)

=\(\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)

Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}\)

=> \(A=\frac{1}{2^2}\left(1-\frac{1}{n}\right)=\frac{1}{4}\left(1-\frac{1}{n}\right)=\frac{1}{4}-\frac{1}{4.n}< \frac{1}{4}\)

Ta có : \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}\)

= \(\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{n^2}\right)\)

< \(\frac{1}{2^2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{\left(n-\right).n}\right)\)

= \(\frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)

= \(\frac{1}{4}.\left(1-\frac{1}{n}\right)\)

<  \(\frac{1}{4}.1=\frac{1}{4}\)

 \(\Rightarrow\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}< \frac{1}{4}.\left(1-\frac{1}{n}\right)< \frac{1}{4}\)

 \(\Rightarrow\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}< \frac{1}{4}\left(đpcm\right)\)

1/

$A=1+2-3-4+5+6-7-8+....+2017+2018-2019-2020+2021+2022$

$=(1+2-3-4)+(5+6-7-8)+...+(2017+2018-2019-2020)+4043$
$=(-4)+(-4)+(-4)+...+(-4)+4043$
Số lần xuất hiện của -4 là: $[(2020-1):1+1]:4=505$

$A=(-4)\times 505+4043=2023$

Câu b có vẻ đề sai. Bạn xem lại nhé.

=A*(4/2+1/2-5/2)

=A*(5/2-5/2)=0