Question

[bảo]

chứng tỏ

1+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9

có chia hết cho 3 không

Answer 1

[Mavis]

\(1+2+2^2+2^3+2^4+2^5+2^6\)\(+2^7+2^8+2^9\)=

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

\(3+2^2\left(1+2\right)+2^4\left(1+2\right)\)\(+2^6\left(1+2\right)+2^8\left(1+2\right)\)=

\(3+2^2.3+2^4.3+2^6.3+2^8.3\)=

\(3.\left(2^2+2^4+2^6+2^8\right)\)

=> biểu thức trên chia hết cho 3(vì trong biểu thức có  thừa số 3)

Answer 2

[Nguyễn Tùng Chi]

1+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9

=(1+2)+2^2(1+2)+2^4(1+2)+2^6(1+2)+2^8(1+2)

=3(1+2^2+2^4+2^6+2^8) 

luôn chia hết cho 3