giúp mình với

Đặt S = B + 2022 + 2023

Số số hạng của B là : ( 2021 - 1 ) : 4 + 1 = 506 ( số )

Tổng B là : ( 2021 + 1 ) . 506 : 2 = 511566

=> S = 511566 + 2022 + 2023

=> S = 515611

Vậy,......

Nhỏ hơn

Ta có 2020/2021 <1

         2021/2022 <1

         2022/2023 <1

         2023/2024 <1

Suy ra A=(2021/2021+2021/2022 +2022/2023 +2023/2024) < (1+1+1+1)= 4

      Vậy A <4

Chúc bạn học tốt

\(\dfrac{2020}{2021}< 1\)

\(\dfrac{2021}{2022}< 1\)

\(\dfrac{2021}{2022}< 1\)

\(\dfrac{2023}{2024}< 1\)

Do đó: A<4

Sửa đề: Tính \(\left(S-P\right)^{2022}\) và \(P=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}\)

Ta có: \(S=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2021}-\frac{1}{2022}\)

\(=\left(1+\frac12+\frac13+\frac14+\cdots+\frac{1}{2021}+\frac{1}{2022}\right)-2\left(\frac12+\frac14+\cdots+\frac{1}{2022}\right)\)

\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{2022}-1-\frac12-\cdots-\frac{1}{1011}\)

\(=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}\)

=P

=>S-P=0

=>\(\left(S-P\right)^{2022}=0\)

\(A=7^{2022}-7^{2021}+7^{2020}-7^{2019}+...+7^2-7\)

\(\Rightarrow7A=7^{2023}-7^{2022}+7^{2021}-...+7^3-7^2\)

\(\Rightarrow8A=A+7A=7^{2022}-7^{2021}+...+7^2-7+7^{2023}-7^{2022}+...+7^3-7^2=7^{2023}-7\)

\(\Rightarrow A=\dfrac{7^{2023}-7}{8}\)

ta có 

\(C=2020\times\left(2021^9+2021^8+...+2021^2+2021^1+1\right)+1\)

\(2020\times\frac{2021^{10}-1}{2021-1}+1=2021^{10}-1+1=2021^{10}\)

Ta có: \(S=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2021}-\frac{1}{2022}\)

\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{2021}+\frac{1}{2022}-2\left(\frac12+\frac14+\cdots+\frac{1}{2022}\right)\)

\(=1+\frac12+\cdots+\frac{1}{2022}-1-\frac12-\cdots-\frac{1}{1011}\)

\(=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}=P\)

=>S-P=0

=>\(\left(S-P\right)^{2022}=0\)

S = \(\left(1+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)

= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)

= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1010}\right)\)

= \(\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2021}\)

S=P nhé

2021 không chia hết cho 2

2022 ⋮ 2

432 ⋮ 2

Vậy A không chia hết cho 2

Lời giải:

$S=1-3+3^2-3^3+...-3^{2021}+3^{2022}$

$3S=3-3^2+3^3-3^4+...-3^{2022}+3^{2023}$

$\Rightarrow S+3S=3^{2023}-1$

$\Rightarrow 4S=3^{2023}-1$

$\Rightarrow 4S-3^{2023}=-1$