Question

Cho A= 3+3²+3³+...+3⁹⁹+3^100.Tìm số tự nhiên n,biết 2A+3=3ⁿ.

Answer 1

A=3+3^2+3^3+..........+3^99+3^100

3A=3^2+3^3+...............+3^100+3^101

=> 3A-A= (3^2+3^3+......+3^100+3^101) - (3+3^2+3^3+........+3^99+3^100)

=> 2A= 3^101 - 3

=>2A+3=3^101

=>3^n=3^101

=> n=101

Answer 2

Ta có:

\(A=3+3^2+3^3+...+3^{99}+3^{100}\)

\(2A=3^2+3^3+3^4+...+3^{100}+3^{101}\)

 

\(2A-A=\left(3^2+3^3+3^4+...+3^{100}+3^{101}\right)-\left(3+3^2+3^3+...+3^{99}+3^{100}\right)\)\(A=3^{101}-3\)

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

\(n=101\)