\(B=3+3^2+3^3+...+3^{2016}\)
\(\Rightarrow3B=3^2+3^3+...+3^{2020}\)
\(\Rightarrow3B-B=3^{2020}-3\)
\(\Rightarrow2B-1=3^{2020}-4\)
( B = 3+3^2+3^3+...+3^2019)
ta có:B = 3 + 3^2+3^3 + ...+ 3^2019
=> 3B = 3^2 + 3^3+3^4 +...+ 3^2020
=> 3B-B = 3^2020 - 3
2B = 3^2020-3
=> 2B -1 = 3^2020 - 3 - 1
2B - 1 = (3^1010)^2 - (3+1)
2B - 1 = (3^1010)^2 - 4 = (3^1010)^2 - 2^2
...
mk chỉ lm đk đến đây thôi! xl bn nha