a) \(S=1+2+3+...+2021\)

\(=\left(2021+1\right).2021:2\)

\(=2043231\)

b) \(P=1+3+5+...+2021\)

\(=\left(2021+1\right).[\left(2021-1\right):2+1]:2\)

\(=2022.1011:2\)

\(=1022121\)

D

D nha

mình đang cần gấp m.n giúp mình nha :33

chia hết cho mấy vậy bạn?

Chọn B

Cảm ơn nha

B/A

\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)

\(=\dfrac{2022\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}=2022\)

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

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

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

Số số hạng của dãy số là:

\(\frac{2021-1}{1}+1=2021-1+1=2021\) (số)

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

=(1+2-3-4)+(5+6-7-8)+...+(2017+2018-2019-2020)+2021

=(-4)+(-4)+...+(-4)+2021

\(=\left(-4\right)\cdot\frac{2020}{4}+2021=-2020+2021=1\)