Cho C = 1 + 3 + 3 mũ 2 + 3 mũ 3 + .... + 3 mũ 11 .Chứng tỏ rằng C chia hết cho 40.
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![](https://rs.olm.vn/images/avt/0.png?1311)
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Ta có: 3^0 + 3^1 + 3^2 + 3^3 + ... + 3^11
= ( 3^0 + 3^1 + 3^2 + 3^3 ) + ... + ( 3^8 + 3^9 + 3^10 + 3^11 )
= 40 + ... + 3^8 . ( 3^0 + 3^1 + 3^2 + 3^3 )
= 40 + ... + 3^8 . 40
= 40 . ( 1 + ... + 3^8 ) \(⋮\)40
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\(1+3+3^2+............+3^{11}\)
\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(3^8+3^9+3^{10}+3^{11}\right)\)
\(=1\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+3^8\left(1+3+3^2+3^3\right)\)
\(=1.40+3^4.40+3^8.40\)
\(=40\left(1+3^4+3^8\right)⋮40\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=1+3+3^2+3^3+...+3^{11}\)
\(C=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{10}+3^{11}\right)\)
\(C=4+3^2.\left(1+3\right)+...+3^{10}.\left(1+3\right)\)
\(C=4+3^2.4+...+3^{10}.4\)
\(C=4.\left(1+3^2+...+3^{10}\right)\)
Vif \(4⋮4=>C⋮4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
chi lam bài 3 nhé
ta thấy:
1+3+32+33+....+311
=( 1 + 3 + 32 + 33 ) + 34 x (1 + 3 + 32 + 33 ) + 38 x ( 1 + 3 + 32 + 33 )
=13 + 34 x 13 + 13 x 38
=>13 x (34 + 38 + 1 )
=>13 x (34 + 38 + 1 ) chia hết cho 13
=>B chia hết cho 13
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b)
=(1+ 3 + 3 mũ 2 + 3 mũ 3) + (1+ 3 + 3 mũ 2 + 3 mũ 3)x34 + (1+ 3 + 3 mũ 2 + 3 mũ 3)x38
=40+40x34+40x38
=40x(34+38+1)
=>40x(34+38+1) chia hết cho 40
=>B chia hết cho 40
k mình nha
\(C=\left(1+3+3^2+3^3\right)+...+\left(3^8+3^9+3^{10}+3^{11}\right)\)
\(C=\left(1+3+3^2+3^3\right)+...+3^8\left(1+3+3^2+3^3\right)\)
\(C=\left(1+3+3^2+3^3\right).\left(1+3^4+3^8\right)\)
\(C=40.\left(1+3^4+3^8\right)\)
Vậy \(C⋮40\)
sửa đề là cho \(C=1+3+3^2+3^3+...+3^{11}\)
Ta có: \(C=1+3+3^2+3^3+3^4+...+3^{11}\)
\(C=4+3^2\left(1+3\right)+3^4\left(1+3\right)+3^6\left(1+3\right)+...+3^{10}\left(1+3\right)\)
\(C=4+3^2.4+3^4.4+3^6.4+...+3^{10}.4\)
\(C=4\left(1+3^2+3^4+3^6+3^8+3^{10}\right)⋮4\left(ĐPCM\right)\)
VẬy C chia hết cho 4