Question

[Phan Trong Nhan]

cho A=75(4^2004+4^2003+4^2002+....+4^2+4+1)+25

    CM:A chia hết cho 100

     AI GIÚP MIK VS Ạ CHIỀU ĐI HOK R  

      THANKS

Answer 1

[๖ۣۜℳเйɦ☂đąйǥ☂ɦọς☂Ŧ๏áйッ]

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

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

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

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

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

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

Thay B vào A ta có

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

\(A=25.3.(\text{​​}\text{​​}\frac{4^{2005}-1}{3})+25\)

\(A=25.(\text{​​}\text{​​}4^{2005}-1)+25\)

\(A=25.(\text{​​}\text{​​}4^{2005}-1+1)\)

\(A=25.\text{​​}\text{​​}4^{2005}\)

Hok tốt !!!!!!!!!

Answer 2

[Trần Đại Nghĩa]

\(A=75\left(4^{2004}+4^{2003}+4^{2002}+...+4^2+4+1\right)+25\)

\(=75\cdot4^{2004}+75\cdot4^{2003}+75\cdot4^{2002}+...+7\cdot4^2+75\cdot4+\left(75+25\right)\)

\(=3\cdot\left(25\cdot4\right)\cdot4^{2003}+3\cdot\left(25\cdot4\right)\cdot4^{2002}+3\cdot\left(25\cdot4\right)\cdot4^{2001}+...+3\cdot\left(25\cdot4\right)\cdot4+3\cdot\left(25\cdot4\right)+25\cdot4\)

\(=3\cdot100\cdot4^{2003}+3\cdot100\cdot4^{2002}+3\cdot100\cdot4^{2001}+...+3\cdot100\cdot4+3\cdot100+100\)

Mà:

\(3\cdot100\cdot4^{2003}⋮100\)

\(3\cdot100\cdot4^{2002}⋮100\)

\(3\cdot100\cdot4^{2001}⋮100\)

\(...\)

\(3\cdot100⋮100\)

\(100⋮100\)

\(\Rightarrow3\cdot100\cdot4^{2003}+3\cdot100\cdot4^{2002}+3\cdot100\cdot4^{2001}+...+3\cdot100\cdot4+3\cdot100+100⋮100\)

\(\Rightarrow A⋮100\left(đpcm\right)\)