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\(4^{2024}-7=4^2.4^{2022}-7=16.\left(4^3\right)^{674}-7=16.64^{674}-7\)
Do \(64\equiv1\left(mod9\right)\Rightarrow64^{674}\equiv1\left(mod9\right)\)
\(\Rightarrow16.64^{674}\equiv16\left(mod9\right)\)
\(\Rightarrow16.64^{674}-7\equiv16-7\left(mod9\right)\)
Mà \(16-7=9⋮9\Rightarrow4^{2024}-7⋮9\)
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\(S=\left(1+4\right)+\left(4^2+4^3\right)+...+\left(4^{98}+4^{99}\right)\\ S=\left(1+4\right)+4^2\left(1+4\right)+...+4^{98}\left(1+4\right)\\ S=\left(1+4\right)\left(1+4^2+...+4^{98}\right)=5\left(1+4^2+...+4^{98}\right)⋮5\)
\(S=\left(1+4\right)+...+4^{98}\left(1+4\right)\)
\(=5\left(1+...+4^{98}\right)⋮5\)
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a: \(D=3^2+3^4+...+3^{120}\)
\(=3\cdot3+3\cdot3^3+...+3\cdot3^{119}\)
\(=3\left(3+3^3+...+3^{119}\right)⋮3\)
b: \(D=3^2+3^4+3^6+...+3^{120}\)
\(=3^2+3^2\cdot3^2+3^2\cdot3^4+...+3^2\cdot3^{118}\)
\(=3^2\left(1+3^2+3^4+...+3^{114}+3^{116}+3^{118}\right)\)
\(=9\cdot\left[\left(1+3^2+3^4\right)+3^6\left(1+3^2+3^4\right)+...+3^{114}\left(1+3^2+3^4\right)\right]\)
\(=9\cdot91\left[1+3^6+...+3^{114}\right]⋮91\)
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Bài 1:
\(2^{49}=\left(2^7\right)^7=128^7;5^{21}=\left(5^3\right)^7=125^7\\ Vì:128^7>125^7\Rightarrow2^{49}>5^{21}\)
Bài 2:
\(a,S=1+3+3^2+3^3+...+3^{99}\\ =\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\\ =40+3^4.40+...+3^{96}.40\\ =40.\left(1+3^4+...+3^{96}\right)⋮40\\ b,S=1+4+4^2+4^3+...+4^{62}\\ =\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{60}.\left(1+4+4^2\right)\\ =21+4^3.21+...+4^{60}.21\\ =21.\left(1+4^3+...+4^{60}\right)⋮21\)
Bài 1 :
\(2^{49}=\left(2^7\right)^7=128^7\)
\(5^{21}=\left(5^3\right)^7=125^7\)
mà \(125^7< 128^7\)
\(\Rightarrow2^{49}>5^{21}\)
Bài 2 :
a) \(S=1+3+3^2+3^3+...3^{99}\)
\(\Rightarrow S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)...+3^{96}\left(1+3+3^2+3^3\right)\)
\(\Rightarrow S=40+40.3^4+...+40.3^{96}\)
\(\Rightarrow S=40\left(1+3^4+...+3^{96}\right)⋮40\)
\(\Rightarrow dpcm\)
b) \(S=1+4+4^2+4^3+...4^{62}\)
\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)
\(\Rightarrow S=21+4^3.21+...4^{60}.21\)
\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)
\(\Rightarrow dpcm\)