Chứng minh S = 5+5^2+5^3+…+5^300 chia hết cho 6 và 31
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Alaia
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Đáp án:
Giải thích các bước giải: $$S=5+5^2+5^3+…+5^{300}\\\Leftrightarrow 5S=5^2+5^3+…+5^{301}\\\Leftrightarrow 5S-S=5^{301}-5=4S\\\Leftrightarrow S=\frac{5^{301}-5}{4}$$
Ta có : $$\frac{5^6-5}{4}\equiv 5(mod\hspace{0,1cm}6)\\\Leftrightarrow 5^6-5\equiv 2(mod\hspace{0,1cm}6)\\\Leftrightarrow 5^6\equiv 1(mod\hspace{0,1cm}6)\\\Leftrightarrow …..$$
Đáp án:
$\begin{array}{l}
S = 5 + {5^2} + {5^3} + … + {5^{300}}\\
= \left( {5 + {5^2}} \right) + \left( {{5^3} + {5^4}} \right) + … + \left( {{5^{299}} + {5^{300}}} \right)\\
= 5\left( {1 + 5} \right) + {5^3}\left( {1 + 5} \right) + … + {5^{299}}\left( {1 + 5} \right)\\
= 5.6 + {5^3}.6 + … + {5^{299}}.6\\
= \left( {5 + {5^3} + … + {5^{299}}} \right).6 \vdots 6\\
S = 5 + {5^2} + {5^3} + … + {5^{300}}\\
= \left( {5 + {5^2} + {5^3}} \right) + \left( {{5^4} + {5^5} + {5^6}} \right) + … + \left( {{5^{298}} + {5^{299}} + {5^{300}}} \right)\\
= 5\left( {1 + 5 + {5^2}} \right) + {5^4}\left( {1 + 5 + {5^2}} \right) + … + {5^{288}}\left( {1 + 5 + {5^2}} \right)\\
= 5.31 + {5^4}.31 + … + {5^{288}}.31\\
= \left( {5 + {5^4} + … + {5^{288}}} \right).31 \vdots 31
\end{array}$