Question

cho A=3+3^2+3^3+.....+3^99. Tìm số tự nhiên n, biết rằng : 2A+3=3^n

Answer 1

=2

Answer 2

\(A=3+3^2+3^3+3^4+....+3^{98}+3^{99}\)

\(\Leftrightarrow3A=3\left(3+3^2+3^3+3^4+....+3^{98}+3^{99}\right)\)

\(\Leftrightarrow3A=3^2+3^3+3^4+3^5+....+3^{98}+3^{99}+3^{100}\)

\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+3^5+....+3^{98}+3^{99}+3^{100}\right)-\left(3+3^2+3^3+3^4+....+3^{98}+3^{99}\right)\)

\(\Leftrightarrow2A=3^{100}-3\)

\(\Leftrightarrow2A+3=3^{100}=3^n\)

\(\Rightarrow n=100\)

Vậy \(n=100\)