S=3+3mũ2+3mũ3+....+3mũ2022
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28 tháng 1 2024
S=3+3^2+3^3+...+3^2022
3S=3.(3+3^2+3^3+...+3^2022)
3S=3^2+3^3+3^4+...+3^2023
⇒3S-S=(3^2+3^3+3^4+...+3^2023)-(3+3^2+3^3+...+3^2022)
⇒2S=3^2023-3
⇒S=3^2023-3 / 2
28 tháng 1 2024
S=3+3^2+3^3+...+3^2022
=>3S=3^2+3^3+3^4+...+3^2023
=>3S-S=(3^2+3^3+3^4+...+3^2023)-(3+3^2+3^3+...+3^2022)
=>2S=3^2023-3
=>S=\(\dfrac{3^{2023}-3}{2}\)
Vậy S=\(\dfrac{3^{2023}-3}{2}\)
LT
0
11 tháng 10 2018
a/
\(3S=3+3^2+3^3+3^4+...+3^{120}\)
\(2S=3S-S=3^{120}-1\Rightarrow S=\frac{3^{120}-1}{2}\)
b/ \(S=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)
\(S=13+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)
\(S=13+3^3.13+...+3^{117}.13=13\left(1+3^3+...+3^{117}\right)\) chia hết cho 13
c/
\(S=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{116}+3^{117}+3^{118}+3^{119}\right)\)
\(S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{116}\left(1+3+3^2+3^3\right)\)
\(S=40+3^4.40+...+3^{116}.40=40\left(1+3^4+...+3^{116}\right)\) chia hết cho 40
NT
6
10 tháng 8 2019
\(A=3^1+3^2+3^3+3^4+...+3^{199}\)
\(3A=3^2+3^3+3^4+3^5+...+3^{200}\)
\(3A-A=\left(3^2+3^3+3^4+...+3^{200}\right)-\left(3^1+3^2+3^3+...+3^{199}\right)\)
\(2A=3^{200}-3^1\)
\(A=\frac{3^{200}-3}{2}\)
=))
XO
10 tháng 8 2019
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 \(3^1+3^2+3^3+..+3^{199}=\frac{3^{200}-1}{2}\)
DB
4
26 tháng 9 2023
\(3^6:3^2+2^3.2^2-3^3.3\)
\(=3^4+2^5-3^4\)
\(=3^4-3^4+2^5\)
\(=0+2^5=2^5\)
NM
26 tháng 9 2023
\(3^6:3^2+2^3.2^2-3^3.3\\ =3^4+2-3^4\\ =\left(3^4-3^4\right)+2\\ =0+2\\ =2.\)
LN
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
1 tháng 10 2018
a)31x32x33x........x3100
=31+2+3+4+...+100
=3(100+1)x(100-1+1):2
=3101x100:2
=35050
Bài b mình không biết làm
NH
1
\(\Rightarrow3S=3^2+3^3+3^4+...+3^{2023}\)
trừ vế với vế ta được :
\(3S-S=3^{2023}-3\)
\(\Rightarrow2S=3^{2023}-3\)
\(\Rightarrow S=\dfrac{3^{2023}-3}{2}\)