Câu hỏi của Đỗ Đức Anh

Question

các bạn ơi cho mình hỏi bài này với :

CM: 2^0+2^1+2^2+2^3+2^4+2^5+....+2^97+2^98+2^99 chia hết cho 31

Nguyễn Thị Ngọc Thơ · Accepted Answer

$2^0+2^1+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}$

$=\left(2^0+2^1+2^2+2^3+2^4\right)+2^5\left(1+2^1+2^2+2^3+2^4\right)$$+...+2^{95}\left(1+2^1+2^2+2^3+2^4\right)$

$=31\left(1+2^5+...+2^{95}\right)$$⋮31$(đpcm)

Câu trả lời:

$2^0+2^1+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}$

$=\left(2^0+2^1+2^2+2^3+2^4\right)+2^5\left(1+2^1+2^2+2^3+2^4\right)$$+...+2^{95}\left(1+2^1+2^2+2^3+2^4\right)$

$=31\left(1+2^5+...+2^{95}\right)$$⋮31$(đpcm)