Question

Cho S=1+3+3²+3³+3⁴+...3⁹⁹

Hãy chứng minh rằng:    S\(⋮\)40

Answer 1

\(S=1+3+3^2+3^3+3^4+...+3^{99}\)

\(S=3^0+3^1+3^2+3^4+...+3^{99}\)

\(\Rightarrow S=3^0.\left(1+3+9+27\right)+...+3^{96}.\left(1+3+9+27\right)\)

\(\Rightarrow S=3^0.40+...+3^{96}.40\)

\(\Rightarrow S=\left(3^0+...+3^{96}\right).40⋮40\)

\(\Rightarrowđpcm\)

Answer 2

S=1+3+3²+...+3⁹⁹

=(1+3+3²+3³)+.....+(3⁹⁶+3⁹⁷+3⁹⁸+3⁹⁹)

=1*(1+3+3²+3³)+....+3⁹⁶*(1+3+3²+3³)

=1*(1+3+9+27)+...+3⁹⁶*(1+3+9+27)

=1*40+....+3⁹⁶*40

=40*(1+...+3⁹⁶) chia hết 40

Đpcm