Cho C 1 3 3 2 3 3 ... 3 11. Chứng minh rằng a, C chia hết cho 13b, C chia hết cho 40
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\(C=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+......+\left(3^9+3^{10}+3^{11}\right)\)
\(C=13.1+3^3.13+......+3^9.13\)
\(C=13.\left(1+3^3+3^6+3^9\right)\)
Chia hết cho 13
\(C=\left(1+3+3^2+3^3\right)+......+\left(3^8+3^9+3^{10}+3^{11}\right)\)
\(C=40.1+40.3^4+40.3^8\)
\(C=40.\left(1+3^4+3^8\right)\)
Chia hết cho 40
Cho A = 1-3+3 mũ 2-3 mũ 3+3 mũ 4-3 mũ 5+.....+3 mũ 98-3 mũ 99 chứng to A chia hết cho 20
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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 3C = 3 + 3^2 + 3^3 + ...3^12
=> 3C - C = (3 + 3^2 + 3^3 + ...3^12) - (1+3+3^2+3^3+....+3^11) = 3^12 - 1 = 531440
hay 2C = 531440 => C = 53144 :2 = 265720
265720 = 20440.13 => C chia hết cho 13 ( vì có thừa số 13)
265720 = 6643.40 => C chia hết cho 40 ( vì có thừa số 40)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : \(C=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^9+3^{10}+3^{11}\right)\)
\(=\left(1+3+3^2\right)+3^3.\left(1+3+3^2\right)+...+3^9.\left(1+3+3^2\right)\)
\(=13+3^3.13+...+3^9.13\)
\(=13.\left(1+3^3+...+3^9\right)⋮13\)
\(\Rightarrow C⋮13\left(\text{đpcm}\right)\)
b) Ta có : \(C=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(3^8+3^9+3^{10}+3^{11}\right)\)
\(=\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^4\right)+3^8.\left(1+3+3^2+3^3\right)\)
\(=40+3^4.40+3^8.40\)
\(=40.\left(1+3^4+3^8\right)⋮40\)
\(\Rightarrow C⋮40\left(\text{đpcm}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b)=3^1+(3^2+3^3+3^4)+(3^5+3^6+3^7)+....+(3^58+3^59+3^60)
=3^1+(3^2.1+3^2.3+3^2.9)+(3^5.1+3^5.3+3^5.9)+......+(3^58.1+3^58.3+3^58.9)
=3^1+3^2.(1+3+9)+3^5.(1+3+9)+.....+3^58.(1+3+9)
=3+3^2.13+3^5.13+.........+3^58.13
=3.13.(3^2+3^5+....+3^58)
vi tich tren co thua so 13 nen tich do chia het cho 13
=
bai1
a) A=(31+32)+(33+34)+...+(359+360)
=(3^1.1+3^1.3)+...+(3^59.1+3^59.2)
=3^1.(1+3)+...+3^59.(1+3)
=3^1.4+....+3^59.4
=4.(3^1+...+3^59)
vi tich tren co thua so 4 nen tich do chia het cho 4