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18 tháng 12 2021

Answer:

Cách 1:

Số số hạng:

\(\left(999-1\right):2+1=500\) số hạng

Tổng:

\(\left(999+1\right).500:2=250000\)

Cách 2:

\(1+3+5+7+...+997+999\)

\(=\left(1+999\right).[\left(999-1\right):2+1]:2\)

\(=1000.500:2\)

\(=250000\)

Ta có: \(S=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2014}\)

\(\Leftrightarrow\dfrac{-1}{7}\cdot S=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+\left(-\dfrac{1}{7}\right)^3+...+\left(-\dfrac{1}{7}\right)^{2015}\)

\(\Leftrightarrow S-\dfrac{-1}{7}\cdot S=\left(-\dfrac{1}{7}\right)^0-\left(-\dfrac{1}{7}\right)^{2015}\)

\(\Leftrightarrow\dfrac{8}{7}\cdot S=1+\dfrac{1}{7^{2015}}\)

\(\Leftrightarrow S=\left(1+\dfrac{1}{7^{2015}}\right):\dfrac{8}{7}=\dfrac{\left(1+\dfrac{1}{7^{2015}}\right)\cdot7}{8}\)

26 tháng 1 2021

Cười cái gì mà cười

17 tháng 9 2016

\(S=\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+...+\left(-\frac{1}{7}\right)^{2017}\)

\(-\frac{1}{7}S=\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2018}\)

\(S-\left(-\frac{1}{7}S\right)=\left[\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+...+\left(-\frac{1}{7}\right)^{2017}\right]-\left[\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2018}\right]\)

\(S+\frac{1}{7}S=\left(-\frac{1}{7}\right)^0-\left(-\frac{1}{7}\right)^{2018}\)

\(\frac{8}{7}S=1+\left(\frac{1}{7}\right)^{2018}\)

\(S=\frac{1+\frac{1}{7^{2018}}}{\frac{8}{7}}=\frac{\left(1+\frac{1}{7^{2018}}\right).7}{8}\)

6 tháng 5 2017

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=\(\dfrac{(\dfrac{-1}{7})^{2017}-1}{\dfrac{-8}{7}}\)