\(B=3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3B=3\left(3+3^2+3^3+...+3^{100}\right)\)
\(\Rightarrow3B=3^2+3^3+3^4+...+3^{101}\)
\(\Rightarrow3B-B=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+3^3+3^{100}\right)\)
\(\Rightarrow2B=3^{101}-3\)
Mà \(2B+3=3^n\)
\(\Rightarrow3^{101}-3+3=3^n\)
\(\Rightarrow3^{101}=3^n\)
\(\Rightarrow n=101\)
Vậy \(n=101\)
Ta có :
\(B\) = \(3\) + \(3^2\) + \(3^3\) + ........ + \(3^{100}\) ( 100 số hạng)
\(3\)\(B\)= \(3^2\) + \(3^3\) + .............+ \(3^{100}\) + \(3^{101}\)
2B = \(3^{101}\) - 3
=> 2B + 3 = \(3^{101}\)
\(3^{101}\) = \(3^n\)
=> n = 101
Vậy n = 101 là giá trị cần tìm
giúp mk nha
