Question

tính M= \(20009\left(2001^9+2001^8+...+2001^1\right)\)

Answer 1

Ta có:

\(M=20009.\left(2001^9+2001^8+....+2001^1\right)\)

\(2001M=20009.\left(2001^{10}+2001^9+....+2001^2\right)\)

\(\Rightarrow2001M-M=20009\left(2001^{10}+...+2001^2\right)-20009\left(2001^9+...+2001^1\right)\)

\(\Rightarrow2000M=20009\left(2001^{10}-2001^1\right)\)

\(\Rightarrow M=\dfrac{20009\left(2001^{10}-2001\right)}{2000}\)

\(\Rightarrow M=10,0045.2001\left(2001^9-1\right)\)

\(\Rightarrow M=20019,0045.\left(2001^9-1\right)\)

Chúc bạn học tốt!!!