Question

Cho B=3+\(^{3^2}\)+\(^{3^3}\)+....+\(3^{2015}\).Hãy tìm số tự nhiên n, biết rằng 2B+3=\(3^n\)

Answer 1

Ta có

\(B=3+3^2+3^3+....+3^{2015}\)

\(3B=3^2+3^3+....+3^{2016}\)

\(\Rightarrow3B-B=\left(3^2+3^3+....+3^{2016}\right)-\left(3+3^2+....+3^{2015}\right)\)

\(\Rightarrow2B=3^{2016}-3\)

\(\Rightarrow2B+3=3^{2016}\)

Answer 2

Ta có:
\(B=3+3^2+...+3^{2015}\)

\(\Rightarrow3B=3^2+3^3+3^4+...+3^{2016}\)

\(\Rightarrow3B-B=\left(3^2+3^3+...+3^{2016}\right)-\left(3+3^2+...+3^{2016}\right)\)

\(\Rightarrow2B=3^{2016}-3\)

Thay 2B vào \(2B+3=3^n\) ta có:

\(3^{2016}-3+3=3^n\)

\(\Rightarrow3^{2016}=3^n\)

\(\Rightarrow n=2016\)

Vậy n = 2016