A= (2+22)+(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)
Vì 3 chia hết cho 3 => 3.(2+23+...+259)  chia hết cho 3
=>A  chia hết cho 3
A= (2+22+23)+...+(258+259+260)
A=2.(1+2+22)+...+258.(1+2+22)
A=2.7+...+258.7
A=7.(2+...+258)
Vì 7  chia hết cho 7 =>7.(2+...+258)  chia hết cho 7

CHIA HẾT CHO 3 :

A= (2+22)+(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)

Vì 3 chia hết cho 3 => 3.(2+23+...+259) chia hết cho 3

=>A chia hết cho 3

dcv

giúp mik với 

\(A=2^0+2^1+2^2+...+2^{59}\)

\(=2^0\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{57}\left(1+2+2^2\right)\)

\(=2^0.7+2^3.7+...+2^{57}.7\)

\(=7\left(2^0+2^3+...+2^{57}\right)⋮7\)

\(S=1+2+2^2+2^3+2^4+...+2^{2011}\)

\(\Rightarrow S=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{2009}\left(1+2+2^2\right)\)

\(\Rightarrow S=7+2^3.7+...+2^{2009}.7\)

\(\Rightarrow S=7\left(1+2^3+...+2^{2009}\right)⋮7\)

\(\Rightarrow dpcm\)

\(S=\left(1+2\right)+...+2^6\left(1+2\right)=3\left(1+...+2^6\right)⋮3\)

\(S=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{95}+2^{96}\right)\\ S=\left(1+2\right)\left(2+2^3+...+2^{95}\right)\\ S=3\left(2+2^3+...+2^{95}\right)⋮3\left(1\right)\\ S=\left(2+2^2\right)+2^3\left(1+2^2+...+2^{93}\right)\\ S=8+8\left(1+2^2+...+2^{93}\right)⋮8\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow S⋮24\)

S=(1+2)+...+2^6(1+2)=3(1+...+2^6)⋮3

S = (1+ 2)+(22 + 23 )+( 24 + 27) + (26 + 25)

S=   3+45+51+51

S=3+3.15+3.17+3.17

S=3.(1+15+17.2): hết 3

tick nha nhanh nhất nè

a: \(2A=2^2+2^3+...+2^{61}\)

=>A=2^61-2

b: \(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+...+2^{55}+2^{58}\right)\) chia hết cho 7(1)

\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)=3\left(2+2^3+...+2^{59}\right)⋮3\left(2\right)\)

Từ (1), (2) suy ra A chia hết cho 21

vì tổng của S chia hết cho 3 nên S chia hết cho 3. có thế cũng hỏi =))

Chúc bạn an toàn