A = 2+22+23+24+...+260
= 2.1+2.2+2.22+2.23+...+2.259
=2.(1+22+23+...259)
=> A chia hết cho 2
A = 2+22+23+24+...+260
=(2+22)+(23+24)+...+(259+260)
=(2.1+2.2)+(23.1+23.2)+...+(259.1+259.2)
=2.(1+2)+23.(1+2)+...+259.(1+2)
=2.3+23.3+...+259.3
=3.(2+23+....+259)
A = 2+22+23+24+...+260
=(2+22+23)+(24+25+26)....+(258+259+260)
=(2.1+2..2+2.22)+(24.1+24.2+24.22)...+(258.1+258.2+258.22)
=2.(1+2+22)+24.(1+2+22)...+258.(1+2+22)
=2.7+24.7...+258.7
=7.(2+24....+258)
=> A chia hết cho 7
A = 2+22+23+24+...+260
=(2+22+23)+(24+25+26)+...+(258+259+260)
=14+(23.2+23.22+23.23)+...+(257.2+257.22+257.23)
=14.1+23.(2+22+23)+...+257.(2+22+23)
=14.1+23.14+...+257.14
=14.(1+23+...+257)
=>A chiaa hết cho 14
Vậy A chia hết cho 2;3;7;14
Mình có câu trả lời ngắn hơn
* A=2+22+23+24+....+260
A=(2+22+23)+(24+25+26)+....+(258+259+260)
A=2.(1+2+4)+24.(1+2+4)+.....+258.(1+2+4)
A= 2.7+24.7+....+258.7
A= 2.7+2.23.7+....+2.257.7
A= 2.7.(23+....+257)
\(\Rightarrow\)A \(⋮\)2,7,14
* A=2+22+23+24+....+260
A=2.1+2.2+23+....+260
A= (2.1+2.2)+(23+24)+....+(259+260)
A= 2.(1+2)+23.(1+2)+.....+259.(1+2)
A= 2.3+23.3+....+259+3
A= 3.(2+23+.....+259)
\(\Rightarrow\)A\(⋮\)3