Câu hỏi của Quách Thanh Nhã

Question

Chứng minh rằng : A= $3+3^2+3^3+....+3^{28}+3^{29}+3^{30}$  chia hết cho 13.

Mashiro Shiina · Accepted Answer

$A=3+3^2+3^3+...+3^{28}+3^{29}+3^{30}$

$A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{29}+3^{30}\right)$

$A=1\left(3+3^2\right)+3^2\left(3+3^2\right)+....+3^{28}\left(3+3^2\right)$

$A=\left(1+3^2+...+3^{28}\right)\left(3+3^2\right)$

$A=13\left(1+3^2+...+3^{28}\right)⋮13\left(đpcm\right)$Câu trả lời: $A=3+3^2+3^3+...+3^{28}+3^{29}+3^{30}$

$A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{29}+3^{30}\right)$

$A=1\left(3+3^2\right)+3^2\left(3+3^2\right)+....+3^{28}\left(3+3^2\right)$

$A=\left(1+3^2+...+3^{28}\right)\left(3+3^2\right)$

$A=13\left(1+3^2+...+3^{28}\right)⋮13\left(đpcm\right)$

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