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)

NHóm để đặt nhân tử có 13 và 40 nhen :3

\(C=1+3+3^2+.......+3^{11}\)

\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+......+3^9\left(1+3+3^2\right)\)

\(=\left(1+3+3^2\right)\left(1+3^3+.....+3^9\right)\)

\(=13.\left(1+3^3+.....+3^9\right)\)

\(\Rightarrow C⋮13\)

C =( 1 + 3 + 3^2) +( 3^3 + 3^4 + 3^5) + ...... + (3^9 + 3^10 + 3^11 )

C = 13.1 + 3^3 .13 + ...... + 3^9 .13

C = 13. (1 + 3^3 + 3^6 + 3^9)

Chia hết cho 13

C = (1 + 3 + 3^2 + 3^3) + ...... + (3^8 + 3^9 + 3^10 + 3^11)

C = 40.1 + 40.3^4 + 40.3^8

C = 40. (1 + 3^4 + 3^8 )

Chia hết cho 40

Vậy......

\(C=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+......+\left(3^9+3^{10}+3^{11}\right)\)

\(C=13.1+3^3.13+......+3^9.13\)

\(C=13.\left(1+3^3+3^6+3^9\right)\)

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

gffjjfhhhfh

* C=(1+3+3²)+(3³+3⁴+3⁵)+...+(3⁹+3¹⁰+3¹¹)

     = 13+3³.(1+3+3²)+...+3⁹.(1+3+3²)

     = 13+3³.13+...+3⁹.13   chia hết cho 13

* Tương tự nhóm 4 số hạng một với nhau.

Chúc bạn học tốt!

1. C chia hết cho 13

C=(1+3+3^2)+(3^3+3^4+3^5)+...+(3^9+3^10+3^11)

  =  13 + 3^3.(1+3+3^2)+...+3^9.(1+3+3^2)

  =  13 + 3^3.13+...+3^9.13

  = 13.(3^3+...+3^9) chia hết cho 13

 (vì 13 chia hết cho 13)

2. C chia hết cho 40

C = 1 + 3 + 32 + 33 + ......+311 

C=30+31+32+...311

C = (30 + 3 + 32 + 33) + (34 + 35 + 36 + 37) + (38 + 39 + 310+ 311)

C = 30(1 + 3 + 32 + 33) + 34(1 + 3 + 32 + 33) + 38(1 + 3 + 32 + 33)

C = 30.40 + 34. 80 + 38. 40

C= 40(30 + 34 + 38) ( chia hết cho 40 vì tích có thừa số 40 

1) \(C=1+3+3^2+3^3+...+3^{11}\)

\(C=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^9+3^{10}+3^{11}\right)\)

\(C=13+3^3\left(1+3+3^2\right)+...+3^9\left(1+3+3^2\right)\)

\(C=13+3^3.13+...+3^9.13\)

\(C=13\left(3^3+3^4+...+3^9\right)\)

Vì \(13\left(3^3+3^4+...+3^9\right)⋮13\)

Vậy \(C=13\left(3^3+3^4+...+3^9\right)⋮13\) (đpcm)

2)

\(C = 1 + 3 + 3^2 + 3^3 + ......+3^{11 }\)

\(C=3^0+3^1+3^2+...3^{11}\)

\(C = (3^0 + 3^1 + 3^2 + 3^3) + (3^4 + 3^5 + 3^6 + 3^7) + (3^8 + 3^9 + 3^{10}+ 3^{11})\)

\(C = 3^0(1 + 3 + 3^2 + 3^3) + 3^4(1 + 3 + 3^2 + 3^3) + 3^8(1 + 3 + 3^2 + 3^3)\)

\(C = 3^0.40 + 3^4.480 + 3^8. 40\)

\(C= 40(3^0 + 3^4 + 3^8)\)

Vì \( 40(3^0 + 3^4 + 3^8) \vdots 40\)

Vậy \(C= 40(3^0 + 3^4 + 3^8) \vdots 40\)

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 : \(C=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(3^8+3^9+3^{10}+3^{11}\right)\)

\(=\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^4\right)+3^8.\left(1+3+3^2+3^3\right)\)

\(=40+3^4.40+3^8.40\)

\(=40.\left(1+3^4+3^8\right)⋮40\)

\(\Rightarrow C⋮40\left(\text{đpcm}\right)\)