Question

Tìm \(x\in Z\) biết: \(1-3+3^2-3^3+...+\left(-3\right)^x=\dfrac{9^{1006}-1}{4}\)

Answer 1

Sửa đề: \(1-3+3^2-3^3+...+\left(-3\right)^x=\dfrac{1-9^{1006}}{4}\)

Đặt \(A=\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+\left(-3\right)^3+...+\left(-3\right)^x\)

\(\Leftrightarrow-3A=\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{x+1}\)

\(\Leftrightarrow-4A=\left(-3\right)^{x+1}-1\)

hay \(A=\dfrac{-\left(-3\right)^{x+1}+1}{4}\)

Theo đề, ta có: \(-\left(-3\right)^{x+1}+1=1-9^{1006}\)

=>x+1=2012

hay x=2011