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a)B=1+3+32+33+....+399
=(1+3)+(32+33)+...+(398+399)
=4+32.4+....+398.4
=4.(1+32+...+398) chia hết cho 4
Vậy B chia hết cho 4
b)B=1+32+33+34+...+399
=(1+3+32+33)+....+(396+397+398+399)
=40+.........+396.40
=40.(1+....+396) chia hết cho 40
Vậy B chia hết cho 40
a)B=(1+3)+(32+33)+...+(398+399)
=(1+3)+32(1+3)+....+398(1+3)
=4+32.4+...+398.4
=4(1+32+...+398) chia hết cho4
câu b bạn vận dụng theo câu a là đc bạn nhóm 4 lại nhé mình hơi lười làm
C/M C\(⋮\)4
\(C=1+3+3^2+...+3^{99}⋮4\)
\(C=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{98}+3^{99}\right)⋮4\)
\(C=\left(1+3\right)+3^2.\left(1+3\right)+...+3^{98}.\left(1+3\right)⋮4\)
\(C=4+3^2.4+...+3^{98}.4⋮4\)
\(C=4.\left(1+3^2+...+3^{98}\right)⋮4\)
C/M C\(⋮\)40
\(C=1+3+3^2+...+3^{99}⋮40\)
\(C=\left(1+3+3^2+3^3\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)⋮40\)
\(C=\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)⋮40\)
\(C=40.1+...+3^{96}.40⋮40\)
\(C=40.\left(1+...+3^{96}\right)⋮40\)
\(C=1+3^1+3^2+...+3^{99}\)
\(=\left(1+3^1\right)+\left(3^2+3^3\right)+...+\left(3^{98}+3^{99}\right)\)
\(=\left(1+3\right)+3^2\left(1+3\right)+...+3^{98}\left(1+3\right)\)
\(=4\left(1+3^2+...+3^{98}\right)\)chia hết cho \(4\).
\(C=1+3^1+3^2+...+3^{99}\)
\(=\left(1+3^1+3^2+3^3\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)
\(=\left(1+3^1+3^2+3^3\right)+...+3^{96}\left(1+3^1+3^2+3^3\right)\)
\(=40\left(1+3^4+...+3^{96}\right)\)chia hết cho \(40\).
\(\text{Ta có:}\)
\(B=1+3+3^2+3^3+3^4+3^5+3^6+3^7+.......+3^{96}+3^{97}+3^{98}+3^{99}\)
\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+.....+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)
\(=40+\left[3^4\left(1+3+3^2+3^3\right)\right]+.....+\left[3^{96}\left(1+3+3^2+3^3\right)\right]\)
\(=40+3^4\cdot40+....+3^{96}\cdot40\)
\(=40\left(1+3^4+....+3^{96}\right)\)
\(\Rightarrow B⋮40\)