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\(S=1+2+2^2+2^3+...+2^{2020}+2^{2021}\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{2020}+2^{2021}\right)\)
\(=3+2^2\left(1+2\right)+...+2^{2020}\left(1+2\right)\)
\(=3+2^2.3+...+2^{2020}.3⋮3\)
VẬY \(S⋮3\)
Trả lời :...........................................
SCSH: (2021 - 1) : 1 = 2020
Tổng: (2021 + 1) : 2 = 1011
Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
k nhé
\(S=1+2+2^2+...+2^9\)
\(\Rightarrow2S=2+2^2+2^3+...+2^{10}\)
\(\Rightarrow S=2^{10}-1\)
Lại có \(5.2^8=\left(2^2+1\right).2^8=2^{10}+2^8\)
Vậy \(S< 5.2^8\)
a, 26 -3(x+1)=14
=> 3(x+1)=26-14
=> 3(x+1)=12
=> x+1=12 : 3
=> x+1=4
=> x=4-1
=> x=3
b, 5x-8=22.23
=> 5x-8=4.8
=> 5x-8=32
=> 5x=32+8
=> 5x=40
=> x=40 : 5
=> x= 8
A.26-3(x+1)=14
3(x+1)=26-14
3(x+1)=12
x+1=12:3
x+1=4
x=4-1=3
B.5x-8=2mux2.2mũ3
5x-8=32
5x=32+8
5x=40
x=40:5
x=8
\(S=\frac{3}{2^0}+\frac{3}{2^1}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(2S=6+\frac{3}{2^0}+\frac{3}{2^1}+...+\frac{3}{2^8}\)
2S-S=6-\(\frac{3}{2^9}\)
S=\(5\frac{509}{512}\)
2x+2x+1+2x+2+2x+3-480=0
2x+2x.2+2x.22+2x.23=0+480
2x.(1+2+22+23)=480
2x.(1+2+4+8)=480
2x.15=480
2x=480:15
2x=32=25
Vậy x =5
nếu sai thì thông cảm nha
\(A=2^2\left(1+2^2\right)+2^6\left(1+2^2\right)+...+2^{18}\left(1+2^2\right)\)
=5(2^2+2^6+...+2^18) chia hết cho 5
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\\ =\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\\ =\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\\ =6+2^2.6+...+2^{98}.6\\ =\left(1+2^2+...+2^{98}\right).6⋮6\left(đpcm\right)\)
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)
\(=6+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)
\(=6\left(1+2^2+....+2^{98}\right)⋮6\)
`S = 2 + 2^2 + 2^3 + 2^4 + ... + 2^2021 + 2^2022`
`= (2 + 2^2) + (2^3 + 2^4) + ... + (2^2021 + 2^2022)`
`= (2 + 2^2) . 1 + (2 + 2^2) . 2^2 + ... + (2 + 2^2) . 2^2020`
`= (2 + 2^2) . (1+2^2 + ... + 2^2020)`
`= 6 . (1+2^2 + ... + 2^2020)`
Do `6 ⋮ 6` nên `6 . (1+2^2 + ... + 2^2020) ⋮ 6 `
`=> S ⋮ 6 (đpcm)`
kệ