Question

Tính tổng : S = \(\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{2004}\)

Answer 1

Answer 2

\(S=\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{2004}\)

\(\left(-3\right)S=\left(-3\right)^1+\left(-3\right)^2+\left(-3\right)^3+...+\left(-3\right)^{2005}\)

\(\left(-3\right)S-S=\left[\left(-3\right)^1+\left(-3\right)^2+\left(-3\right)^3+...+\left(-3\right)^{2005}\right]-\left[\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{2004}\right]\)\(\left(-2\right)S=\left(-3\right)^{2005}-\left(-3\right)^0\)

\(S=\dfrac{\left(-3\right)^{2005}-1}{-2}\)