Câu hỏi của Jimin

Question

tim so du cua A=2^0+2^1+2^2+.............+2^2012 khi chia cho 7

Mashiro Shiina · Accepted Answer

$a=2^0+2^1+2^2+...+2^{2012}$

$=2^0+\left(2^1+2^2+2^3\right)+...+\left(2^{2010}+2^{2011}+2^{2012}\right)$

$=2^0+1\left(2^1+2^2+2^3\right)+...+2^{2009}\left(2+2^2+2^3\right)$

$=2^0+\left(1+...+2^{2009}\right)\left(2^1+2^2+2^3\right)$

$=1+14\left(1+...+2^{2009}\right)$

dư 1Câu trả lời: $a=2^0+2^1+2^2+...+2^{2012}$

$=2^0+\left(2^1+2^2+2^3\right)+...+\left(2^{2010}+2^{2011}+2^{2012}\right)$

$=2^0+1\left(2^1+2^2+2^3\right)+...+2^{2009}\left(2+2^2+2^3\right)$

$=2^0+\left(1+...+2^{2009}\right)\left(2^1+2^2+2^3\right)$

$=1+14\left(1+...+2^{2009}\right)$

dư 1

Violympic toán 7