Dễ thấy:\(64^{12}=\left(4^3\right)^{12}=4^{3.12}=4^{36}\)
Ta có: 4S=\(4\left(4^0+4^1+4^2+4^3+...+4^{35}\right)\)
\(=4^1+4^2+4^3+4^4+...+4^{36}\)
=>4S-S=\(4^{36}-4^0\)
Hay 3S=\(4^{36}-1< 4^{36}=64^{12}\)
Vậy 3S<\(64^{12}\)
Ta có : S=4\(^0\)+4\(^1\)+4\(^2\)+4\(^3\)+ ... + 4\(^{35}\)
Ta thấy : 64\(^{12}\)=(4\(^3\))\(^{12}\)=4\(^{3.12}\)=4\(^{36}\)
Ta sẽ có : 4S=4.(4\(^0\)+4\(^1\)+4\(^2\)+4\(^3\)+ ... + 4\(^{35}\))
=4\(^1\)+4\(^2\)+4\(^3\)+ 4\(^4\)... + 4\(^{36}\)
\(\Rightarrow\)4S-S=4\(^{36}\)-4\(^0\)
Hay : 3S=4\(^{36}\)-1<4\(^{36}\)=64\(^{12}\)
Vậy : 3S<64\(^{12}\)
Dễ thấy:6412=(43)12=43.12=4366412=(43)12=43.12=436
Ta có: 4S=4(40+41+42+43+...+435)4(40+41+42+43+...+435)
=41+42+43+44+...+436=41+42+43+44+...+436
=>4S-S=436−40436−40
Hay 3S=436−1<436=6412436−1<436=6412
Vậy 3S<6412
