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Đặt \(A=1+5+5^2+5^3+...+5^{402}+5^{403}+5^{404}\)
\(\Rightarrow A=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{399}+5^{400}+5^{401}\right)+\left(5^{402}+5^{403}+5^{404}\right)\)
\(\Rightarrow A=31.1+31.5^3+...+31.5^{402}\)
\(\Rightarrow A=31\left(1+5^3+5^6+...+5^{402}\right)\)
\(\Rightarrow A⋮31\left(đpcm\right)\)
\(\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{402}+5^{403}+5^{404}\right)\\ =31+5^3.\left(1+5+5^2\right)+...+5^{402}.\left(1+5+5^2\right)\\ =31+5^3.31+...+5^{402}.31\\ =31.\left(1+5^3+...+5^{402}\right)⋮31\left(DPCM\right)\)
\(B=1+5+5^2+...+5^6+5^7+5^8\)
\(=31+...+5^6\cdot31\)
\(=31\cdot\left(1+...+5^6\right)⋮31\)
ta có:A=(1+5^1+5^2)+(5^3+5^4+5^5)+....+(5^57+5^58+5^59)
=31+31*5^3+....+31*5^57
=(1+5^3+....+5^57)*31
=>a chia hết cho 31
\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=4.\left(3+3^3+...+3^{2009}\right)\)
⇒ \(B\) ⋮ 4
b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)
Ta có: A=1+5+52+…+5403+5404+5405
=>A=(1+5+52)+…+(5403+5404+5405)
=>A=(1+5+52)+…+5403.(1+5+52)
=>A=31+…+5403.31
=>A=(1+…+5403).31 chia hết cho 31
Vậy A chia hết cho 31
Bạn vào câu hỏi tương tự nha !!!