Ta có:

A= 2+2²+2³+...+2²⁰¹⁰+2²⁰¹¹+2²⁰¹²

A=(2+2^2)+(2^3+2^4)+...+(2^2009+2^2010)+(2^2011+2^2012)

A=(2+2^2)+2^2(2+2^2)+...+2^2008(2+2^2)+2^2010(2+2^2)

A=6+2^2x6 + .....+2^2008x6 + 2^2010x6

A=6x(1+2^2+...+2^2008+2^2010) chia hết cho 6 

Vậy A chia hết cho 6

Bạn vào mục câu hỏi tương tự ấy!

 S =(2 + 22) + ( 23 + 24 ) +……..+ ( 22011 + 22012 )
                               = (2 + 22) +26(2 + 22) + ……….22010(2 + 22)
                               =      6       +      22.6   + ………22010.6
                               = 6 ( 1 + 22 + ……+ 22010 )
vậy  chia hết cho 6

ta có A= 2 + 2.2 + 2^3 +.........+2^2012 

   => A = 2 + 4+ 2^3 +.........+ 2^2012

   => A = 6 + 2^3 +...........+ 2^2012

vì 6 chia hết cho 6

=> A : hết cho 6 ( đpcm)

ta có: \(A=2^{ }+2^2+2^3+...+2^{2011}+2^{2012}\)

\(=\)\(\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2011}+2^{2012}\right)\)

\(=\)\(\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{2010}\left(2+2^2\right)\)

\(=\)\(6+2^2.6+...+2^{2010}.6\)

\(=\)\(6.\left(2^2+2^4+...+2^{2010}\right)\)

do \(6⋮6\)

\(\Rightarrow\)\(2^2+2^4+...+2^{2010}⋮6\)

\(\Rightarrow\)\(6.\left(2^2+2^4+...+2^{2010}\right)⋮6\)

\(\Rightarrow\)\(A⋮6\)

vậy A chia hết cho 6

chuk bạn hok tốt