Question

thu gọn rồi tính giá trị
2003x¹⁰⁰+2003x⁹⁹+.....+2003x²+2003x
tại x=2004

Answer 1

A=2003x(1+x+x²+...+x⁹⁸+x⁹⁹)

=> \(\frac{A}{2003x}=1+x+x^2+...+x^{98}+x^{99}\)

=> \(\frac{A.x}{2003x}=x+x^2+...+x^{98}+x^{99}+x^{100}\)=> \(\frac{A}{2003}=x+x^2+...+x^{98}+x^{99}+x^{100}\)

=> \(\frac{A}{2003}-\frac{A}{2003x}=\left(x+x^2+...+x^{98}+x^{99}+x^{100}\right)-\left(1+x+x^2+...+x^{98}+x^{99}\right)\)

=> \(\frac{A\left(x-1\right)}{2003x}=x^{100}-1\)=> \(A=\frac{2003x\left(x^{100}-1\right)}{x-1}\)

Thay x=2004 ta được: \(A=\frac{2003.2004\left(2004^{100}-1\right)}{2004-1}=2004\left(2004^{100}-1\right)\)

Đáp số: \(A=2004\left(2004^{100}-1\right)\)