S=1+2mũ 2+2mũ4+2mũ6+............+2mũ 100
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![](https://rs.olm.vn/images/avt/0.png?1311)
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\(A=2^2\left(1+2^2\right)+2^6\left(1+2^2\right)+...+2^{18}\left(1+2^2\right)\)
=5(2^2+2^6+...+2^18) chia hết cho 5
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 2 + 2² + 2³ + 2⁴ + 2⁵ + ... + 2¹⁰⁰
= 2 + 2².(1 + 2 + 2²) + 2⁵.(1 + 2 + 2²) + ... + 2⁹⁸.(1 + 2 + 2²)
= 2 + 7.2² + 7.2⁵ + ... + 7.2⁹⁸)
= 2 + 7.(2² + 2⁵ + ... + 2⁹⁸)
Vậy số dư khi chia A cho 7 là 2
\(A=2+2^2+2^3+2^4+2^5+...+2^{100}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{97}+2^{98}+2^{99}\right)+2^{100}\)
\(=2\left(1+2+4\right)+2^4\left(1+2+4\right)+...+2^{97}\left(1+2+4\right)+2^{100}\)
\(=7\left(2+2^4+...+2^{97}\right)+2^{100}\)
\(Vì7⋮7=>7\left(2+2^4+..+2^{97}\right)⋮7\)
Ta có:
\(2^3\equiv1\left(mod7\right)\)
\(2^{3.33}\equiv1^{33}\left(mod7\right)\equiv1\left(mod7\right)\)
\(2^{3.33}=2^{99}=>2^{100}=2^{99}.2\equiv1.2\left(mod7\right)\equiv2\left(mod7\right)\)
\(=>2^{100}\) chia \(7\) dư \(2\) mà \(7\left(2+2^4+...+2^{97}\right)⋮7\)
\(=>A\) chia \(7\) dư \(2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\\ =\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\\ =\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\\ =6+2^2.6+...+2^{98}.6\\ =\left(1+2^2+...+2^{98}\right).6⋮6\left(đpcm\right)\)
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)
\(=6+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)
\(=6\left(1+2^2+....+2^{98}\right)⋮6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1+2+2^2+...+2^9\)
\(\Rightarrow2S=2+2^2+2^3+...+2^{10}\)
\(\Rightarrow S=2^{10}-1\)
Lại có \(5.2^8=\left(2^2+1\right).2^8=2^{10}+2^8\)
Vậy \(S< 5.2^8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1+2+2^2+2^3+...+2^{100}\)
\(2S=2+2^2+2^3+2^4+...+2^{101}\)
\(2S-S=\left(2+2^3+..+2^{101}\right)-\left(1+2^2+...+2^{100}\right)\)
\(S=2^{201}-1\)
Ta có
S = 1 + 2 + 22 + 23 + ....+ 2100
2S = 2 + 22 + 23 + 24 + . ....+ 2101
2S-S = ( 2 + 22 + 23 + 24 + . ....+ 2101) - ( 1 + 2 + 22 + 23 + ....+ 2100)
S = 2 + 22 + 23 + 24 + . ....+ 2101 - 1 -2 - 22 - 23 -....- 2100
S = 2101 - 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{3}{2^0}+\frac{3}{2^1}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(2S=6+\frac{3}{2^0}+\frac{3}{2^1}+...+\frac{3}{2^8}\)
2S-S=6-\(\frac{3}{2^9}\)
S=\(5\frac{509}{512}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
S = 2^0 + 2^2 + 2^4 +...+ 2^100
4S = 2^2 + 2^4 + 2^6 + ... + 2^100 + 2^102
4S - S = 2^2 + 2^4 + 2^6 + ... + 2^100 + 2^102 - ( 2^0 + 2^2 + 2^4 +...+ 2^100 )
3S = 2^102 - 1
S = ( 2^102 - 1 ) / 3