Ta có:\(A=1+3+3^2+3^3+3^4+3^5\)
\(\Rightarrow2A=1+3^2+3^3+3^4+3^5+3^6\)
\(\Rightarrow2A+1=1+3^2+3^3+3^4+3^5+3^6\)+1
\(2A+1=2+3^2+3^3+3^4+3^5+3^6\)
Nhớ chọn cho mình nhé,mình sẽ ủng hộ cho
A = 1+ 3\(^2\) + \(3^3\)+ ....+ \(3^5\)
\(\Leftrightarrow\)3A = 3+ \(3^2\)+\(3^3\)+...+\(3^6\)
\(\Leftrightarrow\)3A _ A = (3 + \(3^2\)+....+\(3^6\)) _ (1+ 3 +... +\(3^5\))
\(\Leftrightarrow\)2A = \(3^6\) _ 1
\(\Leftrightarrow\)2A +1 = \(3^6\)( dạng lũy thừa bậc 6 )