Question

Thu gọn biểu thức Q= 3^1 + 3^2 + 3^3 + … + 3^200

Answer 1

\(Q=3^1+3^2+3^3+...+3^{200}\\3\cdot Q=3^2+3^3+3^4+...+3^{201}\\3Q-Q=(3^2+3^3+3^4+...+3^{201})-(3^1+3^2+3^3+...+3^{200})\\2Q=3^{201}-3\\\Rightarrow Q=\dfrac{3^{201}-3}{2}\)

Answer 2

\(Q=3+3^2+...+3^{200}\)

\(3Q=3\cdot\left(3+3^2+...+3^{200}\right)\)

\(3Q=3^2+3^3+...+3^{201}\)

\(3Q-Q=3^2+3^3+...+3^{201}-3-3^2+...-3^{200}\)

\(2Q=3^{201}-3\)

\(Q=\dfrac{3^{201}-3}{2}\)