Question

cho A = 3+3^2+3^3 +..........+3^100

tìm số tự nhiên n biết

2A+3=3n

Answer 1

A = 3¹⁰⁰ + 3

Answer 2

\(A=3+3^2+3^3+...+3^{100}\)

\(3A=3^2+3^3+...+3^{100}+3^{101}\)

\(3A-A=3^2+3^3+...+3^{100}+3^{101}-\left(3+3^2+3^3+...+3^{100}\right)\)

\(2A=3^{101}-3\)

\(2A+3=3n\)

\(\Rightarrow3^{101}-3+3=3n\)

\(\Rightarrow3^{101}=3n\)

\(\Rightarrow n=3^{100}\)

 

 

Answer 3

Ta có A= 3+3^2+3^3+...+3^100

        3A= 3^2+3^3+3^4+...+3^101

   3A-A=(3^2+3^3+3^4+...+3^101)-(3+3^2+3^3+...+3^100)

     2A= 3^101 - 3

Ta lại có 2A+3=3^101-3+3

                    = 3^101

=> 3n=3^101

=> n= 3^101:3

=> n= 3^100

Vậy n= 3^100

Answer 4

3^100

Answer 5

bang 3^100