Câu hỏi của phuong anh

Question

Tính:

K= 2000(2001⁹+2001⁸+...+2001²+2001)+1

✓ ℍɠŞ_ŦƦùM $₦G ✓ · Accepted Answer

$A=2001+2001^2+...+2001^9$

$\Rightarrow2001A=2001^2+2001^3+...+2001^{10}$

$\Rightarrow2001A-A=\left(2001^2+2001^3+...+2001^{10}\right)-\left(2001+2001^2+...+2001^9\right)$$\Rightarrow2000A=2001^{10}-1$

$\Rightarrow A=\frac{2001^{10}-1}{2000}$

$\Rightarrow K=2000.\frac{2001^{10}-1}{2000}+1=2001^{10}-1+1=2001^{10}$

Vậy K=2001¹⁰

Câu trả lời:

$A=2001+2001^2+...+2001^9$

$\Rightarrow2001A=2001^2+2001^3+...+2001^{10}$

$\Rightarrow2001A-A=\left(2001^2+2001^3+...+2001^{10}\right)-\left(2001+2001^2+...+2001^9\right)$$\Rightarrow2000A=2001^{10}-1$

$\Rightarrow A=\frac{2001^{10}-1}{2000}$

$\Rightarrow K=2000.\frac{2001^{10}-1}{2000}+1=2001^{10}-1+1=2001^{10}$

Vậy K=2001¹⁰