Toán (-1/7)^0+(-1/7)^1+(-1/7)^2+….+(-1/7)^2017 31/07/2021 By Emery (-1/7)^0+(-1/7)^1+(-1/7)^2+….+(-1/7)^2017
Đáp án: $\begin{array}{l}A = {\left( { – \dfrac{1}{7}} \right)^0} + {\left( { – \dfrac{1}{7}} \right)^1} + {\left( { – \dfrac{1}{7}} \right)^2} + … + {\left( { – \dfrac{1}{7}} \right)^{2017}}\\ \Rightarrow \left( { – \dfrac{1}{7}} \right).A = {\left( { – \dfrac{1}{7}} \right)^1} + {\left( { – \dfrac{1}{7}} \right)^2} + … + {\left( { – \dfrac{1}{7}} \right)^{2018}}\\ \Rightarrow A – \left( { – \dfrac{1}{7}A} \right) = {\left( { – \dfrac{1}{7}} \right)^{2018}} – {\left( { – \dfrac{1}{7}} \right)^0}\\ \Rightarrow \dfrac{8}{7}A = \dfrac{1}{{{7^{2018}}}} – 1\\ \Rightarrow A = \dfrac{{1 – {7^{2018}}}}{{{{8.7}^{2017}}}}\end{array}$ Trả lời
Đáp án:
⇒A=1−72^0188./7^2017A
Giải thích các bước giải:
Đáp án:
$\begin{array}{l}
A = {\left( { – \dfrac{1}{7}} \right)^0} + {\left( { – \dfrac{1}{7}} \right)^1} + {\left( { – \dfrac{1}{7}} \right)^2} + … + {\left( { – \dfrac{1}{7}} \right)^{2017}}\\
\Rightarrow \left( { – \dfrac{1}{7}} \right).A = {\left( { – \dfrac{1}{7}} \right)^1} + {\left( { – \dfrac{1}{7}} \right)^2} + … + {\left( { – \dfrac{1}{7}} \right)^{2018}}\\
\Rightarrow A – \left( { – \dfrac{1}{7}A} \right) = {\left( { – \dfrac{1}{7}} \right)^{2018}} – {\left( { – \dfrac{1}{7}} \right)^0}\\
\Rightarrow \dfrac{8}{7}A = \dfrac{1}{{{7^{2018}}}} – 1\\
\Rightarrow A = \dfrac{{1 – {7^{2018}}}}{{{{8.7}^{2017}}}}
\end{array}$