Câu hỏi của TÂM TRẦN

Question

$choS=1+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}+2^{11}$chứng tỏ rằng S chia hết cho 3

Cậu Bé Ngu Ngơ · Accepted Answer

$S=1+2+2^2+2^3+...+2^{11}$$=\left(1+2\right)+2^2\left(1+2\right)+...+2^{10}\left(1+2\right)$$=3+3\cdot2^2+3\cdot2^4+3\cdot2^6+3\cdot2^8+3\cdot2^{10}$$=3\left(1+2^2+2^4+2^6+2^8+2^{10}\right)⋮3$Câu trả lời: $S=1+2+2^2+2^3+...+2^{11}$$=\left(1+2\right)+2^2\left(1+2\right)+...+2^{10}\left(1+2\right)$$=3+3\cdot2^2+3\cdot2^4+3\cdot2^6+3\cdot2^8+3\cdot2^{10}$$=3\left(1+2^2+2^4+2^6+2^8+2^{10}\right)⋮3$

Nguyễn Hải Long · Answer

S= (1+2)+22(1+2)+24(1+2)+26(1+2)+28(1+2)+210(1+2)S=3(1+22+24+26+28+210)suy ra S chia hết cho 3Câu trả lời: S= (1+2)+22(1+2)+24(1+2)+26(1+2)+28(1+2)+210(1+2)S=3(1+22+24+26+28+210)suy ra S chia hết cho 3

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