Question

cho s = 3+3^2+3^3+...+3^9   chứng tỏ rằng s chia hết cho 13

Answer 1

giúp tôi với

 

Answer 2

\(S=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\left(3^7+3^8+3^9\right)\\ S=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^7\left(1+3+3^2\right)\\ S=\left(1+3+3^2\right)\left(3+3^4+3^7\right)=13\left(3+3^4+3^7\right)⋮13\)

Answer 3

\(S=3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\\ S=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\left(3^7+3^8+3^9\right)\\ S=3\left(1+3+9\right)+3^4\left(1+3+9\right)+3^7\left(1+3+9\right)\\ S=3.13+3^4.13+3^7.13\\ S=13\left(3+3^4+3^7\right)⋮13\)

Answer 4

S=3+32+33+34+35+36+37+38+39

S=(3+32+33)+(34+35+36)+(37+38+39)

S=3(1+3+9)+34(1+3+9)+37(1+3+9)

S=3.13+34.13+37.13S=13(3+34+37)⋮13