Question

tính tổng:S=1/2+1/2²+...+1/2²⁰¹⁹

^{giúp mị vs}

Answer 1

\(S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}\)

\(\Rightarrow2S=1+\frac{1}{2}+...+\frac{1}{2^{2018}}\)

\(\Rightarrow2S-S=\left(1+\frac{1}{2}+...+\frac{1}{2^{2018}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}\right)\)

\(\Rightarrow S=1-\frac{1}{2^{2019}}\)

Answer 2

Ta có: \(S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}\)

\(\frac{1}{2}S=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}\)

\(\Rightarrow\frac{1}{2}S=S-\frac{1}{2}S=\frac{1}{2}-\frac{1}{2^{2020}}\Rightarrow S=1-\frac{1}{2^{2019}}.\)

Answer 3

\(\frac{1}{2}S=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}\)

\(S-\frac{1}{2}S=\frac{1}{2}-\frac{1}{2^{2020}}\)

\(\Rightarrow\frac{1}{2}S=\frac{1}{2}-\frac{1}{2^{2020}}\)

\(\Rightarrow S=\frac{1}{4}-\frac{1}{2^{2021}}\)

Answer 4

#)Giải :

\(S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}\)

\(\Rightarrow2S=2\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}\right)\)

\(\Rightarrow2S=2+\frac{2}{2}+\frac{2}{2^2}+...+\frac{2}{2^{2019}}\)

\(\Rightarrow2S=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2018}}\)

\(\Rightarrow2S-S=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2018}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}\right)\)

\(S=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2018}}-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{2019}}\)

\(S=2-\frac{1}{2^{2019}}=\frac{2^{2019}-1}{2^{2018}}\)

Vậy : \(S=\frac{2^{2019}-1}{2^{2018}}\)

             #~Will~be~Pens~#