A = 3 + 32 + 33 +...+32019
-> 3A = 3 (3 + 32 + 33 +...+32019)
-> 3A = 32 + 33 + 34 +...+32020
-> 3A - A = (32 + 33 + 34 +...+ 32020) - (3 + 32 + 33 +...+32019)
-> 2A = 32020 - 3
\(\rightarrow A=\frac{3^{2020}-3}{2}\)
Ta có: \(2A+3=3^n\)
\(\Rightarrow2\cdot\frac{3^{2020}-3}{2}+3=3^n\)
\(\Rightarrow3^{2020}-3+3=3^n\)
=> 32020 = 3n => n = 2020
Trl:
\(A=3+3^2+3^3+...+3^{2018}\)
\(3A=3^2+3^3+3^4+...+3^{2017}+3^{2018}\)
\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{100}+3^{101}\right)-\left(3+3^2+3^3+3^4+...+3^{100}\right)\)
\(\Rightarrow2A=3^{101}-3\)
\(\Rightarrow2A+3=3^{101}\)
\(\Rightarrow n=101\)
Vậy n = 101
Hc tốt
A = 3 + 32 + 33 + ... + 32019
3A = 3( 3 + 32 + 33 + ... + 32019 )
= 32 + 33 + 34 + ... + 32020
3A - A = 2A
= ( 32 + 33 + 34 + ... + 32020 ) - ( 3 + 32 + 33 + ... + 32019 )
= 32 + 33 + 34 + ... + 32020 - 3 - 32 - 33 - ... - 32019
= 32020 - 3
2A + 3 = 3n
<=> 32020 - 3 + 3 = 3n
=> 32020 = 3n
=> n = 2020