\(C=1+3+3^2+3^3+...+3^{11}\\ a,C=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+\left(3^6+3^7+3^8\right)+\left(3^9+3^{10}+3^{11}\right)\\ =13+3^3.\left(1+3+3^2\right)+3^6.\left(1+3+3^2\right)+3^9.\left(1+3+3^2\right)\\ =13+3^3.13+3^6.13+3^9.13\\ =13.\left(1+3^3+3^6+3^9\right)⋮13\)

Ý a phải chia hết cho 13 chứ em?

b: C=(1+3+3^2+3^3)+...+3^8(1+3+3^2+3^3)

=40(1+...+3^8) chia hết cho 40

a: C ko chia hết cho 15 nha bạn

\(b,C=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(3^8+3^9+3^{10}+3^{11}\right)\\ =40+3^4.\left(1+3+3^2+3^3\right)+3^8.\left(1+3+3^2+3^3\right)\\ =40.\left(1+3^4+3^8\right)⋮40\)

A=1+3+3^2+3^3+...+3^98+3^99+3^100

A=(1+3+ 3^2)+(3^3+3^4+3^5)+...+(3^98+3^99+3^100)

A=(1+3+3^2)+3^3x(1+3+3^2)+...+3^98x(1+3+3^2)

A=13x3^3x13+...+3^98x13

=> 13x(1+3+3^3+...+3^98)chia hết cho 13

Vậy A chia hết cho 13

câu b đâu bạn ?

cho A = 1 + 3 + 3² + 3³ + ... +  3¹¹

^b) chứng minh A chia hết cho 40

a,  C = 1 + 3 1 + 3 2 + 3 3 + . . . + 3 11

=  1 + 3 1 + 3 2  +  3 3 + 3 4 + 3 5  +...+  3 9 + 3 10 + 3 11

=  1 + 3 1 + 3 2 +  3 3 . 1 + 3 1 + 3 2 + ... +  3 9 1 + 3 1 + 3 2

=  1 + 3 1 + 3 2 . 1 + 3 3 + . . . + 3 9

= 13. 1 + 3 3 + . . . + 3 9 ⋮ 13

b,  C = 1 + 3 1 + 3 2 + 3 3 + . . . + 3 11

=  1 + 3 1 + 3 2 + 3 3 +  3 4 + 3 5 + 3 6 + 3 7 +  3 8 + 3 9 + 3 10 + 3 11

=  1 + 3 1 + 3 2 + 3 3 +  3 4 1 + 3 1 + 3 2 + 3 3 +  3 8 1 + 3 1 + 3 2 + 3 3

=  1 + 3 1 + 3 2 + 3 3 . 1 + 3 4 + 3 8

= 40. 1 + 3 4 + 3 8 ⋮ 40

a: \(A=\left(1+3\right)+...+3^{10}\left(1+3\right)\)

\(=4\left(1+...+3^{10}\right)⋮4\)

C=(1+3+3^2)+3^3(1+3+3^2)+...+3^15(1+3+3^2)

=13(1+3^3+...+3^15) chia hết cho 13