\(M=2+2^2+2^3+...+2^{20}\)
\(M=\left(2+2^2+2^3+2^4\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)
\(M=2\left(1+2+2^2+2^3\right)+...+2^{17}\left(1+2+2^2+2^3\right)\)
\(M=2\cdot15+...+2^{17}\cdot15\)
\(M=15\cdot\left(2+...+2^{17}\right)⋮15\left(đpcm\right)\)
Ta có ;
M = 2 + 22+23+....+220
M = ( 2 + 22+23+24 ) + ....+ ( 217 + 218 + 219 + 220)
M = 2(1 + 2 + 22 + 23)+....+217(1 + 2 + 22 + 23 )
M = 2 . 15 + .... + 217 . 15
Vì 15 chia hết cho 15
Nên 2. 5 + ...+217 . 15
Vậy nên M chia hết cho 15
Bài làm
\(M=2+2^2+2^3+...+2^{20}\)
\(M=\left(2+2^2+2^3+2^4\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)
\(M=2.\left(1+2^2+2^3+2^4\right)+...+2^{17}.\left(1+2+2^2+2^3\right)\)
\(M=2.15+...+2^{17}.15\)
\(M=15.\left(2+...+2^{17}\right)⋮15\left(đpcm\right)\)
# Chúc bạn học tốt #
M=2+22+23+....+220
=(2+22)+(23+24)+....+(219+220)
=2.(1+2)+23.(1+2)+....+219.(1+2)
=2. 3 + 23.3 +....+219.3
= 3.(2+22+....+219) \(⋮\)15
* Vậy M\(⋮\)15.