s=[1+2]+[2+2 mũ 2]+...+[2 mũ 6+2 mũ 7]

s=1 nhân [1+2]+2 nhân [1+2]+...+2 mũ 6 nhân [1+2]

s=[1+2] nhân[1+2+...+2 mũ 6

s=3 nhân [1+2+...+2 mũ 6]

=> s chia hết cho 3

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

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

S=(1+2)+...+2^6(1+2)=3(1+...+2^6)⋮3

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

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

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

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

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

\(\Rightarrow dpcm\)