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a) \(A=5+5^2+5^3+...+5^8\)
\(=\left(5+5^2\right)+5^2\cdot\left(5+5^2\right)+...+5^6\cdot\left(5+5^2\right)\)
\(=\left(5+5^2\right)\cdot\left(1+5^2+...+5^6\right)\)
\(=30\cdot\left(1+5^2+...+5^6\right)\)chia hết cho 30.
b) \(B=3+3^3+3^5+3^7+...+3^{29}\)
\(=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+...+3^{26}\cdot\left(3+3^3+3^5\right)\)
\(=\left(3+3^3+3^5\right)\cdot\left(1+3^6+...+3^{26}\right)\)
\(=273\cdot\left(1+3^6+3^{26}\right)\)chia hết cho 273.
\(A=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{25}+3^{27}+3^{29}\right)\)
\(=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+...+3^{24}\left(3+3^3+3^5\right)\)
\(=273+3^6.273+........+3^{24}.273\)
\(=273\left(1+3^6+......+3^{24}\right)\)chia hết cho 273
Ta có:
273=3+3^3+3^5
A=(3+3^3+3^5)+(3^7+3^9+3^11)+...+(3^25+3^27+3^29)
A=1×(3+3^3+3^5)+3^6×(3+3^3+3^5)+...+3^24×(3+3^3+3^5)
A=1×273+3^6×273+...+3^24×273
A=(1+3^6+...+3^24)×273
Suy ra: A chia hết cho 273
a) \(A=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^6\left(5+5^2\right)=30+5^2.30+...+5^6.30\)
\(=30\left(1+5^2+...+5^6\right)⋮30\Rightarrowđpcm\)
b) \(B=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+...+3^{24}\left(3+3^3+3^5\right)=273+3^6.273+...+3^{24}.273\)
\(=273.\left(1+3^6+...+3^{24}\right)⋮273\Rightarrowđpcm\)
a: \(B=5\left(1+5+5^2+5^3\right)+5^5\left(1+5+5^2+5^3\right)\)
\(=156\cdot5\cdot\left(1+5^4\right)\)
\(=780\left(1+5^4\right)⋮30\)
b: \(B=\left(3+3^3+3^5\right)+...+3^{24}\left(3+3^2+3^5\right)\)
\(=273\cdot\left(1+...+3^{24}\right)⋮273\)
\(B=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{25}+3^{27}+3^{29}\right)\\ B=\left(3+3^3+3^5\right)+3^4\left(3+3^3+3^5\right)+...+3^{24}\left(3+3^3+3^5\right)\\ B=\left(3+3^3+3^5\right)\left(1+3^4+...+3^{24}\right)\\ B=273\left(1+3^4+...+3^{24}\right)⋮273\)
Vậy B là bội 273