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

C=13+3³(1+3+3²)+...+3⁹(1+3+3²)

C=13+3³.13+...+3⁹.13

C=13(1+3³+...+3⁹)

Vì nó có thừa số 13 nên chia hết cho 13 (1+3³+...+3⁹ là STN)

C=(1+3+3²+3³)+(3⁴+3⁵+3⁶+3⁷)+(3⁸+3⁹+3¹⁰+3¹¹)

C=40+3⁴(1+3+3²+3³)+3⁸(1+3+3²+3³)

C=40+3⁴.40+3⁸.40

=40(1+3⁴+3⁸)

=>C chia hết cho 40

a) Ta có: \(A=3+3^3+3^5+...+3^{1991}\)

\(=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{1987}+3^{1989}+3^{1991}\right)\)

\(=3\times\left(1+3^2+3^4\right)+3^7\times\left(1+3^2+3^4\right)+...+3^{1987}\times\left(1+3^2+3^4\right)\)

\(=3\times91+3^7\times91+...+3^{1987}\times91\)

\(=3\times7\times13+3^7\times7\times13+...+3^{1987}\times7\times13\)

\(=13\times\left(3\times7+3^7\times7+...+3^{1987}\times7\right)\)

Vì \(A=13\times\left(3\times7+3^7\times7+...+3^{1987}\times7\right)\)nên A chia hết cho 13.

b) Ta có: \(A=3+3^3+3^5+...+3^{1991}\)

\(=\left(3+3^3+3^5+3^7\right)+...+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\)

\(=3\times\left(1+3^2+3^4+3^6\right)+...+3^{1985}\times\left(1+3^2+3^4+3^6\right)\)

\(=3\times820+...+3^{1985}\times820\)

\(=3\times20\times41+...+3^{1985}\times20\times41\)

\(=41\times\left(3\times20+...+3^{1985}\times20\right)\)

Vì \(A=41\times\left(3\times20+...+3^{1985}\times20\right)\)nên A chia hết cho 41.

C=1+3+3^2+...+3^11

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

C=13+13.3^3+...+13.3^9

C=13(1+3^3+3^6+3^9) chia hết 13

C=1+3+3^2+...+3^11

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

C=40+40.3^4+40.3^8

=40(1+3^4+3^8) chia hết 40

Cho C= 1+3+3²+...+3¹¹

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

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

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

\(=40.\left(1+3^4+3^8\right)\) chia hết cho 40.

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

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

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

\(=13.\left(1+3^3+3^6+3^9\right)\)chia hết cho 13

=> điều phải chứng minh

Rất cảm ơn cậu nhé

Ta có :

A = 13! - 11! = 11! . 12 . 13 - 11! = 11! . (12 . 13 - 1) = 11! . 155 chia hết cho 155

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

Bài 2:(12a + 36b) = (12a + 12 x 3 x b) = 12( a + 3b)chia hết cho 12

Bạn tham khảo 2 link này:

b) https://olm.vn/hoi-dap/detail/104629170538.html

a)https://olm.vn/hoi-dap/detail/8732513603.htm

C=như trên

đến đoạn này mình thấy đề bạn thiếu hay sao ý . đnág nhẽ là C=1+3+3^2+3^3 +..+3^1 ko  nên làm theo cái mình sửa nhá

=> 3C=\(3+3^2+3^3+3^4+...+3^{12}\)

=>3C-C=\(\left(3+3^2+3^3+3^4+...+3^{12}\right)-\left(1+3+3^2+3^3+...+3^{11}\right)\)

=>2C=\(3^{12}-1=531440⋮40\)

=> 2C chia hết cho 40 

=> C cũng chia hết cho 40

a) Ta có:

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

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

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

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

\(=13\left(1+3^3+...+3^9\right)⋮13\)

\(\Leftrightarrow C⋮13\left(đpcm\right)\)

b) Ta có: 

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

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

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

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

\(=40\left(1+3^4+...+3^8\right)⋮40\)

\(\Leftrightarrow C⋮40\left(đpcm\right)\)

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

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

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

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

Vì \(13⋮13\)

Do đó : \(C⋮13\)

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

\(\Rightarrow C=\left(1+3+3^2+3^3\right)+....+\left(3^8+3^9+3^{10}+3^{11}\right)\)

\(\Rightarrow C=40+40.3^4+3^8.40\)

\(\Rightarrow C=40.\left(1+3^4+3^8\right)\)

Vì \(40⋮40\)

Do đó  \(C⋮40\)(đpcm)

a,C1+3+3²)+.....+3⁹,(1+3+3²)

C=13+.....+3⁹.13

C=13(1+.....+3⁹) chia hết cho 13

Vậy C chia hết cho 13

b,C=(1+3+3²+3³)+.....+3⁸(1+3+3²+3³)

   C=40+.....+3⁸+40

    C=40(1+.....+3⁸.40

    C=40(1+.....+3⁸ chia hết cho 40

Vậy C chia hết cho 40

a) C=(1+3+3²)+(3³+3⁴+3⁵)+(3⁶+3⁷+3⁸)+(3⁹+3¹⁰+3¹¹)

    C=1(1+3+3²)+3³(1+3+3²)+...+3⁹(1+3+3²)

     C=13(1+3³+...+3⁹) chia hết cho 13

b)C=(1+3+3²+3³)+(3⁴+3⁵+3⁶+3⁷)+(3⁸+3⁹+3¹⁰+3¹¹)

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

    C=40(1+3⁴+3⁸) chia hết cho 40

C = 1 + 3¹ + 3² + 3³ + ....... + 3¹¹

C = ( 3¹¹ - 3¹ ) : 1¹ = 3⁹

C = 3⁹ - ( 3³ - 3² )

C = 3⁹ - 3¹

C =3⁸