`1+3+3^2+...+3^11`
`=3+1+3^2(1+3)+......+3^10(3+1)`
`=(3+1)(1+3^2+...+3^10)`
`=4(1+3^2+...+3^10) \vdots 4(đpcm)`
`A=(1+3+3^2+3^3+3^4)+....+(....+3^11)`
`=40+3^4 .40 + 3^8 .40`
`=40.(1+3^4+3^8)`
`=4.10 (1+3^4+3^8) ⋮ 4`.
Giải:
A=1+3+32+...+311
A=1.(1+3)+32.(1+3)+34.(1+3)+...+310.(1+3)
A=(1+3).(1+32+34+...+311)
A=4.(1+32+34+...+310) ⋮ 4 (đpcm)
Chúc bạn học tốt!
a) Ta có: \(A=1+3+3^2+...+3^{11}\)
\(=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{10}+3^{11}\right)\)
\(=4+3^2\cdot4+...+3^{10}\cdot4\)
\(=4\left(1+3^2+...+3^{10}\right)⋮4\)(đpcm)