a)
\(B=3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3B=3\left(3+3^2+3^3+...+3^{100}\right)\)
\(\Rightarrow3B=3^2+3^3+3^4+...+3^{101}\)
\(\Rightarrow3B-B=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+3^3+3^{100}\right)\)
\(\Rightarrow2B=3^{101}-3\)
Mà \(2B+3=3^n\)
\(\Rightarrow3^{101}-3+3=3^n\)
\(\Rightarrow3^{101}=3^n\)
\(\Rightarrow n=101\)
Vậy \(n=101\)
a)
B = 3 + 32 + 33 + ... + 3100
3B = 32 + 33 + 34 + ... + 3101
3B - B = 3101 - 3
⇒ 2B = 3101 - 3
⇒ 2B + 3 = 3101 - 3 + 3
⇒ 3n = 3101
⇒ n = 101
Vậy n = 101