Question

Cho S=1+3+3²+3³+3³+...+3⁹⁹

Hãy chứng minh: S\(⋮\) 4

Answer 1

S=1+3+3²+3³+3³+...+3⁹⁹

=(1+3)+(3²+3³)+...+(3⁹⁸+3⁹⁹)

=1*(1+3)+3²(1+3)+...+3⁹⁸(1+3)

=1*4+3²*4+...+3⁹⁸*4

=4*(1+3²+...3⁹⁸) chia hết 4

 

Answer 2

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

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

\(S=3^0.\left(1+3\right)+3^2.\left(1+3\right)+...+3^{98}.\left(1+3\right)\)

\(S=3^0.4+3^2.4+...+3^{98}.4\)

\(S=\left(3^0+3^2+...+3^{98}\right).4⋮4\)

\(\Rightarrowđpcm\)

Answer 3

S=1+3+3²+3³+3³+...+3⁹⁹

=(1+3)+(3²+3³)+...+(3⁹⁸+3⁹⁹)

=1*(1+3)+3²(1+3)+...+3⁹⁸(1+3)

=1*4+3²*4+...+3⁹⁸*4

=4*(1+3²+...3⁹⁸) chia hết 4

Answer 4

khó quá