\(M=75.4\left(4^{2020}+4^{2019}+...+4+1\right)+75+25=\)

\(=300.\left(4^{2020}+4^{2019}+...+4+1\right)+100=\)

\(=100\left[3.\left(4^{2020}+4^{2019}+...+4+1\right)+1\right]⋮100\)

đề thiếu rồi nek bạn

tui lớp 8 ko bt làm :)

hảo hán

Đặt \(A=75\left(4^{2023}+4^{2022}+...+4^2+4+1\right)+25\)

Đặt \(B=4^{2023}+4^{2022}+...+4^2+4+1\)

=>\(4B=4^{2024}+4^{2023}+...+4^3+4^2+4\)

=>\(4B-B=4^{2024}+4^{2023}+...+4^3+4^2+4-4^{2023}-4^{2022}-...-4^2-4-1\)

=>\(3B=4^{2024}-1\)

=>\(B=\dfrac{4^{2024}-1}{3}\)

\(A=75\left(4^{2023}+4^{2022}+...+4^2+4+1\right)+25\)

\(=75\cdot\dfrac{4^{2024}-1}{3}+25\)

\(=25\cdot\left(4^{2024}-1\right)+25\)

\(=25\cdot4^{2024}\)

\(=25\cdot4\cdot4^{2023}=100\cdot4^{2023}⋮100\)

Chắc đặt nhầm lớp rồi

Ta có :\(B=4^{2004}+4^{2003}+...+4^2+4+1\)

\(4B=\left(4^{2004}+4^{2003}+...+4^2+4+1\right).4\)

\(4B=4^{2005}+4^{2004}+...+4^3+4^2+4\)

\(4B-B=\left(4^{2005}+4^{2004}+...+4^3+4^2+4\right)\)\(-\left(4^{2004}+4^{2003}+...+4+1\right)\)

\(3B=\left(4^{2005}-1\right)\)\(\Rightarrow\frac{4^{2005}-1}{3}\)

\(\Rightarrow A=75.\frac{4^{2005}-1}{3}+25\)

\(\Rightarrow A=25.\left(4^{2005}-1\right)+25\)

\(\Rightarrow A=25.\left(4^{2005}-1+1\right)\)

\(\Rightarrow A=25.4.4^{2004}\)

\(\Rightarrow A=100.4^{2004}\)

Mà 100 chia hết 100 nên \(100.4^{2004}\) chia hết cho 100

B=4^0 + 4^1 +...+ 4^2004

4B=4^1+4^2+...+4^2005

3B=4^2004-4^0

B=(4^2004-4^0):3

Thay B vào  ta có :

A=75.(4^2004-4^0):3+25

A=25.(4^2004-4^0)+25

A=25.4^2004

A=100.4^2003

Vậy A chia hết cho 100

CẶC

A=\(75.\left(4^{2004}+4^{2003}+4^2+4+1\right)+25\)

A=\(75.\left(4^{2005}-1\right):3+25\)

A=\(25.\left(4^{2005}-1+1\right)\)

A=\(25.4^{2005}⋮100\)

Nhớ tick cho mình nhé!