Sửa đề: \(S=1+3+3^2+3^3+\cdots+3^{2021}+3^{2022}\)

Ta có: \(S=1+3+3^2+3^3+\cdots+3^{2021}+3^{2022}\)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5+3^6\right)+\left(3^7+3^8+3^9+3^{10}\right)+\cdots+\left(3^{2019}+3^{2020}+3^{2021}+3^{2022}\right)\)

\(=13+3^3\left(1+3+3^2+3^3\right)+3^7\left(1+3+3^2+3^3\right)+...+3^{2019}\left(1+3+3^2+3^3\right)\)

\(=13+40\left(3^3+3^7+\cdots+3^{2019}\right)=3+10+10\cdot4\cdot\left(3^3+3^7+\cdots+3^{2019}\right)\)

=>S chia 10 dư 3

=>S có tận cùng là 3

A = (-1)(-1)^2(-1)^3...(-1)^2019

A = (-1)^1+2+3+...+2019

A = (-1)^2039190

A = 1

S = 1.2.3 + 2.3.4 + 3.4.5 + ... + 2018.2019.2020

4S = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + .... + 2018.2019.2020.4

4S = 1.2.3.4 + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) + ... + 2018.2019.2020.(2021 - 2017)

4S = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 2018.2019.2020.2021 - 2017.2018.2019

4S = 2018.2019.2020.2021

S = 2018.2019.2020.2021 : 4 = ...

cảm ơn bạn nhiều nhé

Ta có: \(P=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2019}{3^{2019}}-\frac{2020}{3^{2020}}\)

=>\(3P=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2019}{3^{2018}}-\frac{2020}{3^{2019}}\)

=>3P+P=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2019}{3^{2018}}-\frac{2020}{3^{2019}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{2019}{3^{2019}}-\frac{2020}{3^{2020}}\)

=>\(4P=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2019}}-\frac{2020}{3^{2020}}\)

Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2019}}\)

=>3A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2018}}\)

=>3A+A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2018}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2019}}\)

=>4A=-1\(-\frac{1}{3^{2019}}\)

=>\(4A=\frac{-3^{2019}-1}{3^{2019}}\)

=>\(A=\frac{-3^{2019}-1}{4\cdot3^{2019}}\)

Ta có: \(4P=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2019}}-\frac{2020}{3^{2020}}\)

\(=1+\frac{-3^{2019}-1}{4\cdot3^{2019}}-\frac{2020}{3^{2020}}=1+\frac{-3^{2020}-3-8080}{4\cdot3^{2020}}=1-\frac14-\frac{8083}{4\cdot3^{2020}}<\frac34\)

=>\(P<\frac{3}{16}\) (ĐPCM)

S = 1 - 2 + 3 - 4 +...+ 2019 - 2020

= ( 1 - 2 ) + ( 3 - 4 ) +...+ ( 2019 - 2020 )

= ( -1 ) + ( -1 ) +...+ ( -1 )

Có số số hạng ( -1 ) là : ( 2019 - 1 ) : 1 + 1 = 2019

=> S = ( -1 ) x 2019 = ( -2019 )

1. 
S = 1-2+3-4+...+2019-2020
S = (1-2)+(3-4)+...+(2019-2020)
S = (-1) + (-1) +...+ (-1)
S = (-1) . 2020 : 2 = -1010
 

2.
(2x-1)(y+2) = 3
\(\Rightarrow\left(2x-1\right);\left(y+2\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Ta có bảng : 
 

Vậy \(\left(x;y\right)\in\left\{\left(1;1\right);\left(0;-5\right);\left(2;-1\right);\left(-1;-3\right)\right\}\)

thiếu đề :(

S=1+(2-3)+(-4+5)+(6-7)+(-8+9)+...+(-2020+2021)
S=1-1+1-1+1+...+1
S=1+0+0+...+0
S=1

\(S=1+2-3-4+...+2017+2018-2019-2020+2021\\ S=\left(1+2-3-4\right)+...+\left(2017+2018-2019-2020\right)+2021\\ S=\left(-4\right)+\left(-4\right)+\left(-4\right)+...+-4+2021\\ S=505.\left(-4\right)+2021\\ S=-2020+2021\\ S=1\)