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\(A=\left(1+7\right)+...+7^{2020}\left(1+7\right)=8\left(1+...+7^{2020}\right)⋮8\)
\(A = (1 + 7) +...+7^2\)\(^0\)\(^2\)\(^0\) \((1 + 7) = 8 (1+...+7^2\)\(^0\)\(^2\)\(^0\)\() \) ⋮\(8\)
ko chia hết.Vì 1+2+3+.......+13 \(⋮\) 1+2+....+13 mà 14 ko\(⋮\) cho 1+2+.......+13
a) \(M=2+2^2+2^3+...+2^{100}\)
\(\Rightarrow M=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(\Rightarrow M=2\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(\Rightarrow M=2.31+...+2^{96}.31\)
\(\Rightarrow M=\left(2+...+2^{96}\right).31⋮31\)
\(\Rightarrow M⋮31\)
b) \(M=2+2^2+2^3+...+2^{100}\)
\(\Rightarrow2M=2^2+2^3+2^4+...+2^{101}\)
\(\Rightarrow2M-M=\left(2^2+2^3+2^4+...+2^{101}\right)-\left(2+2^2+2^3+...+2^{100}\right)\)
\(\Rightarrow M=2^{101}-2\)
a) M = 2 + 22 + 23 + ... + 2100
= (2+22+23+24+25) + (26+27+28+29+210) + ... + (296+297+298+299+2100)
= 2(1+2+22+23+24) + 26(1+2+22+23+24) + ... + 296(1+2+22+23+24)
= 31(2+26+...+296) \(⋮\) 31
b) M = 2 + 22 + ... + 2100
=> 2M = 22 + 23 + ... + 2101
=> 2M - M = 2101 - 2
=> M = 2101 - 2
Do 36 chia hết cho 4
72 chia hết cho 4
Nên 36 + 72 chia hết cho 72
a: \(=2^2\left(1+2\right)+2^4\left(1+2\right)=3\left(2^2+2^4\right)⋮3\)
b: \(=4^{20}\left(1+4\right)+4^{22}\left(1+4\right)=5\left(4^{20}+4^{22}\right)⋮5\)
c: \(A=\left(1+4+4^2\right)+...+4^{96}\left(1+4+4^2\right)\)
\(=21\left(1+...+4^{96}\right)⋮21\)
d: \(B=7\left(1+7\right)+7^3\left(1+7\right)+...+7^{35}\left(1+7\right)\)
\(=8\left(7+7^3+...+7^{35}\right)⋮8\)
\(B=7\left(1+7+7^2\right)+...+7^{34}\left(1+7+7^2\right)\)
\(=57\left(7+...+7^{34}\right)\) chia hếtcho 3 và 19
M=(7^0+7^1)+(7^2+7^3)+....+(7^68+7^69)
M=8+7^2(1+7)+...+7^68.(1+7)
M=8+7^2.8+...+7^68.8
8.(1+7^2+...+7^68) Chia hết cho 4
tick cho mình nhé đúng rồi đấy