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

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

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

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

S = 1 - \(\frac{1}{2^{10}}\)

\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}\)

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

=> \(S=1-\frac{1}{2^{10}}\)

Study well ! >_<

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

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

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

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

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

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

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

\(S=2S-S\)

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

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

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

\(S=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2017}\)

\(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2017.2018}\)

\(\frac{1}{2}S=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(\frac{1}{2}S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(\frac{1}{2}S=\frac{1}{2}-\frac{1}{2018}\)

\(\frac{1}{2}S=\frac{504}{1009}\)

=> \(S=\frac{1008}{1009}\)

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

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

=>\(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)

=>\(S=1-\frac{1}{2^9}=\frac{511}{512}\)

Vậy \(S=\frac{511}{512}\)

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

\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^8}\)

\(\Rightarrow2S-S=1-\frac{1}{2^9}\)

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

bạn nào nhanh mik cho

\(S=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2017}\)

\(S=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2035153}\)

\(S=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{4070306}\)

\(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{2017.2018}\)

\(S=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2017.2018}\right)\)

\(S=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)

\(S=2.\left(\frac{1}{2}-\frac{1}{2018}\right)=2.\frac{504}{1009}=\frac{1008}{1009}\)

Vậy \(S=\frac{1008}{1009}\)

\(S=\frac{1008}{1009}\)

\(S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{100}\)

\(\Rightarrow2S=2+1+\frac{1}{2}+\frac{1}{2^2}...+\frac{1}{99}\)

\(2S-S=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)

\(\Leftrightarrow2S-S=S=2-\frac{1}{2^{100}}=\frac{2^{101}}{2^{100}}-\frac{1}{2^{100}}=\frac{2^{101}-1}{2^{100}}\)