Câu hỏi của NDH DJ

Question

(-3/2)^2+(-3/2)^3 +(-3/2)^4+(-3/2)^5+...+(-3/2)^100

Nguyễn Lê Phước Thịnh · Accepted Answer

$A=\left(-\dfrac{3}{2}\right)^2+\left(-\dfrac{3}{2}\right)^3+...+\left(-\dfrac{3}{2}\right)^{100}$=>$\dfrac{-3}{2}\cdot A=\left(-\dfrac{3}{2}\right)^3+\left(-\dfrac{3}{2}\right)^4+...+\left(-\dfrac{3}{2}\right)^{101}$=>$\dfrac{-5}{2}\cdot A=\left(-\dfrac{3}{2}\right)^{101}-\left(-\dfrac{3}{2}\right)^2$$\Leftrightarrow A=\left[\left(-\dfrac{3}{2}\right)^{101}-\left(-\dfrac{3}{2}\right)^2\right]\cdot\dfrac{-2}{5}$Câu trả lời: $A=\left(-\dfrac{3}{2}\right)^2+\left(-\dfrac{3}{2}\right)^3+...+\left(-\dfrac{3}{2}\right)^{100}$=>$\dfrac{-3}{2}\cdot A=\left(-\dfrac{3}{2}\right)^3+\left(-\dfrac{3}{2}\right)^4+...+\left(-\dfrac{3}{2}\right)^{101}$=>$\dfrac{-5}{2}\cdot A=\left(-\dfrac{3}{2}\right)^{101}-\left(-\dfrac{3}{2}\right)^2$$\Leftrightarrow A=\left[\left(-\dfrac{3}{2}\right)^{101}-\left(-\dfrac{3}{2}\right)^2\right]\cdot\dfrac{-2}{5}$

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