Question

cho biểu thức A=1+5+5²+5³+   ............................+5²⁰¹⁷

tìm số tự nhiên n biết 4a+1=5ⁿ⁺¹

 

Answer 1

\(A=1+5+5^2+...+5^{2017}\)

\(5A=5.\left(1+5+...+5^{2017}\right)\)

\(5A=5+5^2+5^3+...+5^{2018}\)

\(5A-A=5+5^2+...+5^{2018}-1-5-5^2-...-5^{2017}\)

\(4A=5^{2018}-1\)

Thay \(4A=5^{2018}-1\)vào 4 + 1 = 5ⁿ⁺¹, ta có:

\(5^{2018}-1+1=5^{n+1}\)

\(\Rightarrow5^{2018}=5^{n+1}\Rightarrow n+1=2018\Rightarrow n=2017\)