Question

 tính tổng  

Answer 1

Ta có : 

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

\(=>2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)

\(=>2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)

\(=>S=6-\frac{3}{2^9}=6-\frac{3}{512}=\frac{3072}{512}-\frac{3}{512}=\frac{3069}{512}\)

Answer 2

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

\(\Rightarrow2S=6+3+\frac{3}{2^2}+...+\frac{3}{2^8}\)

\(\Rightarrow2S-S=\left(6+3+\frac{3}{2^2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)

\(\Rightarrow S=6-\frac{3}{2^9}=6-\frac{3}{512}=5\frac{509}{512}\)

Answer 3

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

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

Đặt A = \(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\)

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

\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(1+\frac{1}{2}+....+\frac{1}{2^9}\right)\)

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

Thay A = \(2-\frac{1}{2^9}\) vào S ta có:

S = \(3.\left(2-\frac{1}{2^9}\right)=3.2-3\cdot\frac{1}{2^9}=6-\frac{3}{2^9}\)