Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = 1 + 3 + {3^2} + ..... + {3^{50}}\\
= 1 + \left( {3 + {3^2} + .... + {3^{50}}} \right)\\
= 1 + 3.\left( {1 + 3 + {3^2} + .... + {3^{49}}} \right)\\
3 \vdots 3 \Rightarrow 3.\left( {1 + 3 + {3^2} + ..... + {3^{49}}} \right) \vdots 3
\end{array}\)
Do đó, A chia 3 dư 1.
\(\begin{array}{l}
b,\\
A = 1 + 3 + {3^2} + {3^3} + ..... + {3^{49}} + {3^{50}}\\
= \left( {1 + 3} \right) + \left( {{3^2} + {3^3}} \right) + ....... + \left( {{3^{48}} + {3^{49}}} \right) + {3^{50}}\\
= 4 + {3^2}.\left( {1 + 3} \right) + ..... + {3^{48}}\left( {1 + 3} \right) + {3^{50}}\\
= 4 + {3^2}.4 + ..... + {3^{48}}.4 + {\left( {{3^2}} \right)^{25}}\\
= 4.\left( {1 + {3^2} + .... + {3^{48}}} \right) + {9^{25}}\\
4 \vdots 4 \Rightarrow 4.\left( {1 + {3^2} + ..... + {3^{48}}} \right) \vdots 4
\end{array}\)
Ta có:
9 chia 4 dư 1 nên \({9^{25}}\) chia 4 dư 1.
Vậy A chia 4 dư 1.
