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Ta có: \(S=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2014}\)
\(\Leftrightarrow\dfrac{-1}{7}\cdot S=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+\left(-\dfrac{1}{7}\right)^3+...+\left(-\dfrac{1}{7}\right)^{2015}\)
\(\Leftrightarrow S-\dfrac{-1}{7}\cdot S=\left(-\dfrac{1}{7}\right)^0-\left(-\dfrac{1}{7}\right)^{2015}\)
\(\Leftrightarrow\dfrac{8}{7}\cdot S=1+\dfrac{1}{7^{2015}}\)
\(\Leftrightarrow S=\left(1+\dfrac{1}{7^{2015}}\right):\dfrac{8}{7}=\dfrac{\left(1+\dfrac{1}{7^{2015}}\right)\cdot7}{8}\)
\(S=\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+...+\left(-\frac{1}{7}\right)^{2017}\)
\(-\frac{1}{7}S=\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2018}\)
\(S-\left(-\frac{1}{7}S\right)=\left[\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+...+\left(-\frac{1}{7}\right)^{2017}\right]-\left[\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2018}\right]\)
\(S+\frac{1}{7}S=\left(-\frac{1}{7}\right)^0-\left(-\frac{1}{7}\right)^{2018}\)
\(\frac{8}{7}S=1+\left(\frac{1}{7}\right)^{2018}\)
\(S=\frac{1+\frac{1}{7^{2018}}}{\frac{8}{7}}=\frac{\left(1+\frac{1}{7^{2018}}\right).7}{8}\)
S=(\(\dfrac{-1}{7}\))0+(\(\dfrac{-1}{7}\))1+...+(\(\dfrac{-1}{7}\))2016
\(\Rightarrow\)\(\dfrac{-1}{7}S\)=(\(\dfrac{-1}{7}\))1+(\(\dfrac{-1}{7}\))2+...+(\(\dfrac{-1}{7}\))2017
\(\Rightarrow\)\(\dfrac{-1}{7}S\)-\(S\)=\([\) (\(\dfrac{-1}{7}\))1+(\(\dfrac{-1}{7}\))2+...+
(\(\dfrac{-1}{7}\))2017 \(]\)-\([\)(\(\dfrac{-1}{7}\))0+(\(\dfrac{-1}{7}\))1+...+
(\(\dfrac{-1}{7}\))2016\(]\)
=(\(\dfrac{-1}{7}\))1+(\(\dfrac{-1}{7}\))2+...+(\(\dfrac{-1}{7}\))2017-
(\(\dfrac{-1}{7}\))0-(\(\dfrac{-1}{7}\))1-...-(\(\dfrac{-1}{7}\))2016
\(\dfrac{-8}{7}S\)=(\(\dfrac{-1}{7}\))2017-1
S=\(\dfrac{(\dfrac{-1}{7})^{2017}-1}{\dfrac{-8}{7}}\)