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

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

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

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

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

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

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

\(\Leftrightarrow\frac{1}{2}D=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

\(\Leftrightarrow\frac{1}{2}D-D=\frac{1}{2^{10}}-1\)

\(\Leftrightarrow D=\frac{\frac{1}{2^{10}}-1}{-\frac{1}{2}}\)

Vậy \(3.D=3.\left(\frac{\frac{1}{2^{10}}-1}{-\frac{1}{2}}\right)=3.\frac{1023}{512}=\frac{3069}{512}\)

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

=\(\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^{10}}\)

=> \(S-\frac{1}{2}S=\left(3+\frac{3}{2}+...+\frac{3}{2^9}\right)-\left(\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^{10}}\right)\)

=> \(\frac{1}{2}S=3-\frac{3}{2^{10}}\)

=>\(S=\left(3-\frac{3}{2^{10}}\right).2=6-\frac{6}{2^{10}}=6-\frac{3}{2^9}\)

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

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

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

\(\Rightarrow2A=2+1+\frac{1}{2}+...+\frac{1}{2^8}\)

\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

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

Mà \(S=3.A\)

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

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

Chúc bạn học tốt !!! 

Thanks bạn

2S=\(\frac{6}{2}+\frac{6}{2^2}+....+\frac{6}{2^9}\)

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

S=3+3+3+3+3+3+3+3+3

S=27

ta có : 

2.S=\(\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^{10}}\)

2.S-S=\(\left(\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^{10}}\right)-\left(\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)

S=\(\frac{3}{2^{10}}-\frac{3}{2}\)

Ta có : 

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

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

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

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

\(S=\frac{2^{10}.3-3}{2^9}\)

Vậy \(S=\frac{2^{10}.3-3}{2^9}\)

vận dụng 3S lên

xong tìm S nha bn ok

tại k có thời gian nên chỉ giúp thế thôi

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

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

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

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

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

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

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

\(\Rightarrow S=3\left(2-\frac{1}{2^9}\right)=\frac{3.\left(2^{10}-1\right)}{2^9}\)