Question

Tính tổng:\(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)

Answer 1

Đặt \(A=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)

Nên \(2.A=6+3+\frac{3}{2}+....+\frac{3}{2^8}\)

Suy ra \(2.A-A=6-\frac{3}{2^9}\Rightarrow A=6-\frac{3}{2^9}\)

Vậy giá trị biểu thức là : \(6-\frac{3}{2^9}\)

Answer 2

đặt \(A=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)

\(A=3.\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

đặt \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)( 1 )

\(2B=2+1+\frac{1}{2}+...+\frac{1}{2^8}\)( 2 )

Lấy ( 2 ) - ( 1 ) ta được :

\(B=2-\frac{1}{2^9}\)

\(\Rightarrow A=3.\left(2-\frac{1}{2^9}\right)\)

\(\Rightarrow A=6-\frac{3}{2^9}\)