3 mũ 1+3mũ2+3mũ3+3mũ4+...3mũ199
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Những câu hỏi liên quan
LN
1
NH
1
LP
1
17 tháng 9
Ta có: \(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)
\(=3\left(1-3+3^2-3^3+\cdots+3^{22}-3^{23}\right)\) ⋮3
Ta có: \(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)
\(=3\left(1-3\right)+3^3\left(1-3\right)+\cdots+3^{23}\left(1-3\right)\)
\(=3\cdot\left(-2\right)+3^3\cdot\left(-2\right)+\cdots+3^{23}\left(-2\right)\)
\(=-2\left(3+3^3+\cdots+3^{23}\right)\)
\(=-2\left\lbrack\left(3+3^3\right)+\left(3^5+3^7\right)+\cdots+\left(3^{21}+3^{23}\right)\right\rbrack\)
\(=-2\cdot\left\lbrack3\cdot\left(1+3^2\right)+3^5\left(1+3^2\right)+\cdots+3^{21}\left(1+3^2\right)\right\rbrack\)
\(=-2\cdot10\cdot\left(3+3^5+\cdots+3^{21}\right)=-20\cdot\left(3+3^5+\cdots+3^{21}\right)\) ⋮20
Ta có: \(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)
\(=\left(3-3^2+3^3\right)-\left(3^4-3^5+3^6\right)+\cdots-\left(3^{22}-3^{23}+3^{24}\right)\)
\(=3\left(1-3+3^2\right)-3^4\left(1-3+3^2\right)+\cdots+3^{22}\left(1-3+3^2\right)\)
\(=7\cdot\left(3-3^4+\cdots-3^{22}\right)\) ⋮7
Ta có: C⋮20
C⋮7
C⋮3
mà ƯCLN(20;3;7)=1
nên C⋮20*3*7
=>C⋮420
TH
1
22 tháng 9 2020
Đặt \(D=3-3^2+3^3-3^4+...+3^9-3^{10}+3^{11}\)
=> \(3D=3^2-3^3+3^4-3^5+...+3^{10}-3^{11}+3^{12}\)
Cộng vế 2 BT trên ta được:
\(D+3D=\left(3-3^2+...+3^{11}\right)+\left(3^2-3^3+...+3^{12}\right)\)
\(\Leftrightarrow4D=3^{12}+3\)
\(\Rightarrow D=\frac{3^{12}+3}{4}\)
LT
0
17 tháng 9
Ta có: \(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)
\(=3\left(1-3+3^2-3^3+\cdots+3^{22}-3^{23}\right)\) ⋮3
Ta có: \(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)
\(=3\left(1-3\right)+3^3\left(1-3\right)+\cdots+3^{23}\left(1-3\right)\)
\(=3\cdot\left(-2\right)+3^3\cdot\left(-2\right)+\cdots+3^{23}\left(-2\right)\)
\(=-2\left(3+3^3+\cdots+3^{23}\right)\)
\(=-2\left\lbrack\left(3+3^3\right)+\left(3^5+3^7\right)+\cdots+\left(3^{21}+3^{23}\right)\right\rbrack\)
\(=-2\cdot\left\lbrack3\cdot\left(1+3^2\right)+3^5\left(1+3^2\right)+\cdots+3^{21}\left(1+3^2\right)\right\rbrack\)
\(=-2\cdot10\cdot\left(3+3^5+\cdots+3^{21}\right)=-20\cdot\left(3+3^5+\cdots+3^{21}\right)\) ⋮20
Ta có: \(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)
\(=\left(3-3^2+3^3\right)-\left(3^4-3^5+3^6\right)+\cdots-\left(3^{22}-3^{23}+3^{24}\right)\)
\(=3\left(1-3+3^2\right)-3^4\left(1-3+3^2\right)+\cdots+3^{22}\left(1-3+3^2\right)\)
\(=7\cdot\left(3-3^4+\cdots-3^{22}\right)\) ⋮7
Ta có: C⋮20
C⋮7
C⋮3
mà ƯCLN(20;3;7)=1
nên C⋮20*3*7
=>C⋮420
NP
4
28 tháng 3 2019
\(Q=1+3+3^2+3^3+3^4+...+3^{11}\)
\(3Q=3+3^2+3^3+3^4+3^5+...+3^{12}\)
\(3Q-Q=\left(3+3^2+3^3+3^4+3^5+...+3^{12}\right)-\left(1+3+3^2+3^3+3^4+...+3^{11}\right)\)
\(2Q=3^{12}-1\)
\(Q=\frac{3^{12}-1}{2}\)
28 tháng 9
Ta có: \(S=3+3^2+3^3+\cdots+3^{2024}\)
\(=\left(3+3^2\right)+\left(3^3+3^4+3^5\right)+\left(3^6+3^7+3^8\right)+\cdots+\left(3^{2022}+3^{2023}+3^{2024}\right)\)
\(=12+3^3\left(1+3+3^2\right)+3^6\left(1+3+3^2\right)+\cdots+3^{2022}\left(1+3+3^2\right)\)
\(=12+13\left(3^3+3^6+\cdots+3^{2022}\right)\)
=>S không chia hết cho 13
\(A=3^1+3^2+3^3+3^4+...+3^{199}\)
\(3A=3^2+3^3+3^4+3^5+...+3^{200}\)
\(3A-A=\left(3^2+3^3+3^4+...+3^{200}\right)-\left(3^1+3^2+3^3+...+3^{199}\right)\)
\(2A=3^{200}-3^1\)
\(A=\frac{3^{200}-3}{2}\)
=))
Đặt \(A=3^1+3^2+3^3+...+3^{199}\)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{200}\)
Lấy 3A trừ A theo vế ta có :
\(3A-A=\left(3^2+3^3+3^4+..+3^{200}\right)-\left(3^1+3^2+3^3+..+3^{199}\right)\)
\(2A=3^{200}-1\)
\(A=\frac{3^{200}-1}{2}\)
Vậy \(3^1+3^2+3^3+..+3^{199}=\frac{3^{200}-1}{2}\)