Ta có : 3 + \(\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^9}\)
= 3. ( \(\frac{1}{2^2}+\frac{1}{2^3}...+\frac{1}{2^9}\) )
Đặt A = \(\frac{1}{2^2}+\frac{1}{2^3}...+\frac{1}{2^9}\)
2A = \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
2A - A = \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\) - ( \(\frac{1}{2^2}+\frac{1}{2^3}...+\frac{1}{2^9}\) )
A = \(\frac{1}{2}-\frac{1}{2^9}\)
=> 3. ( \(\frac{1}{2}-\frac{1}{2^9}\) )
=> 3. ( \(\frac{1}{2}-\frac{1}{512}\) )
=> 3. \(\frac{255}{512}\)
=> \(1\frac{253}{512}\)
Đặt \(A=3+\frac{3}{2}+\frac{3}{2^2}+..+\frac{3}{2^9}\)
\(2A=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(2A-A=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+...+\frac{3}{2^9}\right)\)
\(A=6-\frac{3}{2^9}\)
\(A=\frac{3072-3}{512}\)
\(A=\frac{3069}{512}\)
Ủng hộ mk nha !!! ^_^
Đặt \(A=3+\frac{3}{2}+\frac{3}{2^2}+..+\frac{3}{2^9}\)
\(2A=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(2A-A=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+...+\frac{3}{2^9}\right)\)
\(A=6-\frac{3}{2^9}\)
\(A=\frac{3072-3}{512}\)
\(A=\frac{3069}{512}\)