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\(A=1+3+3^2+...+3^{2007}\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{2008}\)
\(\Rightarrow3A-A=\left(3+3^2+3^3+...+3^{2008}\right)-\left(1+3+3^2+...+3^{2007}\right)\)
\(\Rightarrow2A=3+3^2+3^3+...+3^{2008}-1-3-3^2-...-3^{2007}\)
\(\Rightarrow2A=3^{2008}-1\)
\(\Rightarrow2A+1=3^{2008}\)
\(A=1+3+3^2+...+3^{2007}\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{2008}\)
\(\Rightarrow3A-A=\left(3+3^2+3^3+...+3^{2008}\right)-\left(1+3+3^2+...+3^{2007}\right)\)
\(\Rightarrow2A=3+3^2+3^3+...+3^{2008}-1-3-3^2-...-3^{2007}\)
\(\Rightarrow2A=3^{2008}-1\)
\(\Rightarrow2A+1=3^{2008}\)
Nhớ k cho mk nha!!!
a: \(4^5\cdot8^7=2^{10}\cdot2^{21}=2^{31}\)
b: \(125^5\cdot25^3=5^{15}\cdot5^6=5^{21}\)
3A=\(3+3^2+3^3+...+3^{11}\)
3A-A=(\(3+3^2+3^3+...+3^{11}\))-(\(1+3+3^2+...+3^{10}\))
2A=\(3^{11}-1\)
2A+1=\(3^{11}\)
Ta có: \(A=2+2^2+2^3+...+2^{100}\)
\(2A=2^2+2^3+2^4+...+2^{101}\)
\(2A-A=2^{101}-2\)
Hay \(A=2^{101}-2\)
Vậy \(A=2^{101}-2\)
_Học tốt_
3A=3+32+33+....+32008
2A=(3+32+....+32008)-(1+3+...+32007)=32008-1