Question

Cho S = 4^0 + 4^1 + 4^2 + 4^3 + ... + 4^35

Hãy so sánh 3S với 64^12

Answer 1

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

\(\Leftrightarrow4S=4+4^2+4^3+........+4^{36}\)

\(\Leftrightarrow4S-S=\left(4+4^2+......+4^{36}\right)-\left(1+4+4^2+......+4^{35}\right)\)

\(\Leftrightarrow3S=4^{36}-1\)

\(\Leftrightarrow3S+1=4^{36}=\left(4^3\right)^9=64^9< 64^{12}\)

\(\Leftrightarrow3S+1< 64^{12}\)