Question

Chứng tỏ A = \(75\times\left(4^{2004}+4^{2003}+.....+4^2+4+1\right)+25\) là số chia hết cho 100

Answer 1

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

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

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

\(\Leftrightarrow4B-B=\left(4^{2005}+4^{2004}+......+4^2+4\right)-\left(4^{2004}+4^{2003}+.....+4+1\right)\)

\(\Leftrightarrow3B=4^{2005}-1\)

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

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

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

\(\Leftrightarrow A=25.4.4^{2003}\)

\(\Leftrightarrow A=100.4^{2003}⋮100\left(đpcm\right)\)