Đáp án:
n=1009
Giải thích các bước giải:
\(\begin{array}{l}A = {5^1} + 5_.^2... + {5^{2018}}\\5A = {5^2} + {5^3} + ... + {5^{2018}} + {5^{2019}}\\ \Rightarrow 5A - A = \left( {{5^2} + {5^3} + ... + {5^{2018}} + {5^{2019}}} \right) - \left( {{5^1} + 5_.^2... + {5^{2018}}} \right)\\ \Leftrightarrow \,\,\,\,\,\,\,4A\,\,\, = \,{5^{2019}} - 5\,\,\,\,\,\,\,\,\left( 1 \right)\\ma\,\,\,\,\,\,4A + 5 = {5^{2n + 1}} + 1 \Rightarrow 4A = {5^{2n + 1}} - 5\,\,\,\,\,\,\,\,\left( 2 \right)\\Tu\,\,\,\left( 1 \right)\,\,\,va\,\,\,\left( 2 \right)\,\,\,suy\,\,ra:\,\\{5^{2n + 1}} - 5 = {5^{2019}} - 5\\ \Rightarrow 2n + 1 = 2019\\ \Rightarrow 2n = 2018\\ \Rightarrow n = 1009\end{array}\)
