Question

Cho S = 4^0 + 4^1 + 4^2 + 4^3 +...+ 4^35Hãy so sánh 3S với 64^12    
  

Answer 1

\(4^0+4^1+4^2+4^3+...+4^{35}\\ 4S=4^1+4^2+4^3+4^4+...+4^{36}\\ 4S-S=\left(4^1+4^2+4^3+4^4+...+4^{36}\right)-\left(4^0+4^1+4^2+4^3+...+4^{35}\right)\\ 3S=4^{36}-1=64^{12}-1\\ Vì64^{12}-1< 64^{12}\\ \Rightarrow3S< 64^{12}\)

Answer 2

Ta có: \(64^{12}=\left(4^3\right)^{12}=4^{36}\)

\(S=4^0+4^1+...+4^{34}+4^{35}\)

\(\Rightarrow4S=4^1+4^2+...+4^{35}+4^{36}\)

\(\Rightarrow4S-S=4^{36}-4^0\)

\(\Rightarrow3S=4^{36}-1< 4^{36}\)

Vậy \(3S< 64^{12}\)