\(C=1+3+3^2+3^3+...+3^{11}\\ a,C=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+\left(3^6+3^7+3^8\right)+\left(3^9+3^{10}+3^{11}\right)\\ =13+3^3.\left(1+3+3^2\right)+3^6.\left(1+3+3^2\right)+3^9.\left(1+3+3^2\right)\\ =13+3^3.13+3^6.13+3^9.13\\ =13.\left(1+3^3+3^6+3^9\right)⋮13\)
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) C không thể chia hết cho 15
b) Ta có: \(C=1+3+3^2+3^3+...+3^{11}\)
\(C=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(3^8+3^9+3^{10}+3^{11}\right)\)
\(C=\left(1+3+9+27\right)+3^4\left(1+3+9+27\right)+3^8\left(1+3+9+27\right)\)
\(C=40+3^4\cdot40+3^8\cdot40\)
\(C=40\cdot\left(1+3^4+3^8\right)\)
Mà: \(40\cdot\left(1+3^4+3^8\right)\) ⋮ 40
\(\Rightarrow C\) ⋮ 40
Phần a sai đề nha phải là CM C chia hết cho 13
\(a,C=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+3^6\left(1+3+3^2\right)+3^9\left(1+3+3^2\right)\\ C=\left(1+3+3^2\right)\left(1+3^3+3^6+3^9\right)\\ C=13\left(1+3^3+3^6+3^9\right)⋮13\)
\(b,C=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(+3^8+3^9+3^{10}+3^{11}\right)\\ C=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+3^8\left(1+3+3^2+3^3\right)\\ C=\left(1+3+3^2+3^3\right)\left(1+3+3^8\right)\\ C=40\left(1+3^4+3^8\right)⋮40\)