Question

cho a= 1+2002+2002^2 +2002^3 +...+ 2002^99 ; b= 2002^100.  CTR:   b > 2001.a

Answer 1

2002a = \(2002+2002^2+...+2002^{100}\)

=> 2002a -a = \(2002^{100}-1

Answer 2

  Ta có \(B=2002^{100}\)

Ta có \(A=1+2002+2002^2+...+2002^{99}\)

         \(\Rightarrow2002A=2002+2002^3+...+2002^{100}\)

\(\Rightarrow2002A-A=\left(2002+2002^2+2002^3+...+2002^{100}\right)-\left(1+2002+2002^2+...+2002^{99}\right)\)

\(\Rightarrow2002A-A=2002+2002^2+2002^3+...+2002^{100}-1-2002-2002^2-...-2002^{99}\)

\(2001A=2002^{100}-1\)

                           vÌ 2002¹⁰⁰-1<2002¹⁰⁰ nên => A<B

                              ĐÚNG NHÉ

Answer 3

A < B nha bạn

k tui nha

thanks