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

\(\Leftrightarrow4B=4^{2005}+4^{2004}+...+4^3+4^2+4\)

\(\Leftrightarrow B=\dfrac{4^{2005}-1}{3}\)

\(A=75\cdot B+25\)

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

\(=25\cdot4^{2005}=100\cdot4^{2004}⋮100\)

Đặt B = 4²⁰⁰⁴ + 4²⁰⁰³ + 4²⁰⁰² + 4²⁰⁰¹ + ... + 4² + 4 + 1 (có 2005 số; 2005 : 2 dư 1)

B = (4²⁰⁰⁴ + 4²⁰⁰³) + (4²⁰⁰² + 4²⁰⁰¹) + ... + (4² + 4) + 1

B = 4²⁰⁰³.(4 + 1) + 4²⁰⁰¹.(4 + 1) + ... + 4.(4 + 1) + 1

B = 4²⁰⁰³.5 + 4²⁰⁰¹.5 + ... + 4.5 + 1

B = 5.(4²⁰⁰³ + 4²⁰⁰¹ + ... + 4) + 1

=> B = 5 x k + 1 ( k thuộc N*; k chia hết cho 4)

=> A = 75 x (5 x k + 1) + 25

=> A = 75 x 5 x k + 75 + 25

=> A = ...00 + 100

=> A = ..00 chia hết cho 100

A = 75.4^2004 + ... + 75.4 + 75 + 25 
= 25.3.4^2004 + ... + 25.3.4 + 100 
= 100.3.4^2003 + ... + 100.3 + 100 
=> A chia hết cho 100

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

\(\Rightarrow A=75.4^{2004}+75.4^{2003}+....+75.4+75.1+25\)

\(\Rightarrow A=\left(75.4\right).4^{2003}+....+300+100\)

\(\Rightarrow A=300.4^{2003}+.....+300+100\) chia hết cho 100

=> ĐPCM

B=4^2004+4^2003+...+4^2+4+1
Xét 4B = 4^2005+4^2004+...+4^2+4
=> 4B-B = (4^2005+4^2004+...4^3+4^2+4) - (4^2004+4^2003+...+4^2+4+1)
=> 3B = 4^2005 - 1 => B = (4^2005 - 1)/3
=> A = 75 (4^2005 - 1)/3 +25
= 25 (4^2005 -1) +25
= 25 x 4 ^ 2005
= 25 x 4 x 4 ^ 2004 = 100 x4 ^ 2004

Xin lỗi nha ở ngoài ngoặc còn có +25

A=75(4²⁰⁰⁴+4²⁰⁰³+..+4+1)+25

   =75(4²⁰⁰⁴+4²⁰⁰³+..+4)+75+25

   =3.25.(4²⁰⁰⁴+4²⁰⁰³+...+4)+100

   =3.25.4(4²⁰⁰³+4²⁰⁰²+...+1)+100

   =3.100(4²⁰⁰³+4²⁰⁰²+..+1)+100\(⋮\)100

=> A\(⋮\)100

Đúng thì k nha

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