Câu hỏi của Lê Phan Bảo Nguyên

Question

Chứng minh:                                      A = 2 + 22 + 23 + 24  +25 .......+ 299 + 2100   Chia hết cho 62

Mashiro Shiina · Accepted Answer

$A=2+2^2+2^3+2^4+2^5+...+2^{99}+2^{100}$

$A=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)$

$A=1\left(2+2^2+2^3+2^4+2^5\right)+2^5\left(2+2^2+2^3+2^4+2^5\right)+...+2^{95}\left(2+2^2+2^3+2^4+2^5\right)$

$A=\left(2+2^2+2^3+2^4+2^5\right)\left(1+2^5+...+2^{95}\right)$

$A=62\left(1+2^5+...+2^{95}\right)⋮62\left(đpcm\right)$Câu trả lời: $A=2+2^2+2^3+2^4+2^5+...+2^{99}+2^{100}$

$A=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)$

$A=1\left(2+2^2+2^3+2^4+2^5\right)+2^5\left(2+2^2+2^3+2^4+2^5\right)+...+2^{95}\left(2+2^2+2^3+2^4+2^5\right)$

$A=\left(2+2^2+2^3+2^4+2^5\right)\left(1+2^5+...+2^{95}\right)$

$A=62\left(1+2^5+...+2^{95}\right)⋮62\left(đpcm\right)$

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