Question

CMR biểu thức A=75.(4^2017+4^2016+..+4^2+5)+25 chia hết cho 4^2018

Answer 1

Đặt \(A_1=\left(1+4+4^2+...+4^{2016}+4^{2017}\right)\)

Ta có: \(A_1=\left(1+4+4^2+...+4^{2016}+4^{2017}\right)\)

   \(\Leftrightarrow4A_1=4+4^2+4^3+...+4^{2017}+4^{2018}\)

Lấy \(4A_1-A_1\)ta có:

      \(4A_1-A_1=\left(4+4^2+4^3+...+4^{2017}+4^{2018}\right)-\left(1+4+4^2+...+4^{2016}+4^{2017}\right)\)

\(\Leftrightarrow3A_1=4^{2018}-1\)

\(\Leftrightarrow A_1=\frac{4^{2018}-1}{3}\)

Thay \(A_1=\frac{4^{2018}-1}{3}\)vào biểu thức A, ta có: 

         \(A=75.\left(\frac{4^{2018}-1}{3}\right)+25\)

  \(\Leftrightarrow A=25.\left(4^{2018}-1\right)+25\)

  \(\Leftrightarrow A=25.4^{2018}⋮4^{2018}\)

Vậy \(A⋮4^{2018}\)

chúc bn hok tốt

Answer 2

thanks bro

Answer 3

you're welcome