tham khảo nhá Câu hỏi của Nguyễn Khắc Thành - Toán lớp 6 - Học toán với OnlineMath

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

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

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

2 3/7+(2/9-1 3/7)-5/3:3/4=1

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

      \(=9+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^8}\)

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

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

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

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

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

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

\(S=6-\frac{3}{512}\)

\(S=\frac{3069}{512}\)

Vậy \(S=\frac{3069}{512}\)

S=3+3/2+3/2²+...+3/2⁹

S=3/1.2+3/2.3+...+3/8.9

S=3-3/2+3/2-3/3+3/3-3/4+......+3/8-3/9

S=3-3/9=27/9-3/9=24/9=8/3

\(=\frac{8}{3}\)

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

\(\Rightarrow\frac{1}{2}S=\frac{3}{2}+\frac{3}{2^2}+\frac{3}{2^3}+.....+\frac{3}{2^{10}}\)

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

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

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

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

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

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

\(\Rightarrow2A-A=A=1-\frac{1}{2^9}\)

Do đó \(S=3\left(1-\frac{1}{2^9}\right)=3\left(1-\frac{1}{512}\right)=3-\frac{3}{512}=\frac{1533}{512}\)

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

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

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

\(\frac{2S}{3}\)=2+1+\(\frac{1}{2}\)+.............+\(\frac{1}{2^8}\)    (2)

trừ cả 2 vế của (1) và (2) ta được

\(\frac{S}{3}\)=2 -\(\frac{1}{2^9}\)

\(\frac{S}{3}\)=\(\frac{2^{10}-1}{2^9}\)

s=\(\frac{2^{10}-1}{3\cdot2^9}\)