Question

C = 1+\(3^2\)+\(3^4\)+\(3^6\)+....+\(3^{20}\)

Answer 1

\(C=1+3^2+3^4+3^6+...+3^{20}\)

\(\Rightarrow3^2C=3^2+3^4+3^6+3^8+...+3^{22}\)

\(\Rightarrow3^2C-C=3^{22}-1\)

\(\Leftrightarrow8C=3^{22}-1\)

\(\Leftrightarrow C=\dfrac{3^{22}-1}{8}\)

Answer 2

\(C=1+3^2+3^4+...+3^{20}\)

\(\Rightarrow9C=3^2+3^4+3^6+...+3^{22}\)

\(\Rightarrow9C-C=\left(3^2+3^4+3^6+...+3^{22}\right)-\left(1+3^2+3^4+...+3^{20}\right)\)

\(\Rightarrow8C=3^{22}-1\)

\(\Rightarrow C=\dfrac{3^{22}-1}{8}\)

Answer 3

322−18

Answer 4

Ta có: \(C=1+3^2+3^4+3^6+...+3^{20}\)

\(\Leftrightarrow9\cdot C=3^2+3^4+3^6+...+3^{22}\)

\(\Leftrightarrow8\cdot C=3^{22}-1\)

\(\Leftrightarrow C=\dfrac{3^{22}-1}{8}\)