ta có: A = 2 + 22 + 23 + 24 + ...+ 22004 ( có 2004 số hạng)
A = (2+22) + (23 + 24) + ...+ (22003 + 22004) ( có 1002 nhóm số hạng)
A = 2.(1+2) + 23.(1+2) + ...+ 22003.(1+2)
A = 2.3 + 23.3 + ...+ 22003.3
A = 3.(2+23 + ...+ 22003) chia hết cho 3 (đpcm)
chia hết cho 5;7 chứng minh tương tự nha bn!
A=2+2^2^3+...............+2^2004
A=2.(1+2)+2^3.(1+2)+...................+2^2003.(1+2)
A=2.3+2^3.3+........................2^2003.3
suy ra A chia hết cho 3
câu chia hết cho 5 thì
2.(1+2+4+8)+............+2^2001.(1+2+4+8)
=2.15+...............+2^2001.15
suy ra A chia hết cho 5
\(A=2+2^2+2^3+2^4+....+2^{2004}.\)
\(=2.\left(1+2\right)+2^3.\left(1+2\right)+....+2^{2003}.\left(1+2\right)\)
\(=2.3+2^3.3+...+2^{2003}.3\)
\(=3.\left(2+2^3+....+2^{2003}\right)\)
\(\Rightarrow A⋮3\)
\(A=2+2^2+2^3+2^4+....+2^{2004}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+.....+\left(2^{2001}+2^{2002}+2^{2003}+2^{2004}\right)\)
\(=\left[2.\left(1+2+2^2+2^3\right)\right]+\left[2^5.\left(1+2+2^2+2^3\right)\right]....+\left[2^{2001}.\left(1+2+2^2+2^3\right)\right]\)
\(=2.15+2^5.15+....+2^{2001}.15\)
\(=15.\left(2+2^5+....+2^{2001}\right)\)
\(=3.5.\left(2+2^5+...+2^{2001}\right)\)
\(\Rightarrow A⋮5\)
\(A=2+2^2+2^3+2^4+....+2^{2004}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+....+\left(2^{2002}+2^{2003}+2^{2004}\right)\)
\(=\left[2.\left(1+2+2^2\right)\right]+\left[2^4.\left(1+2+2^2\right)\right]+....+\left[2^{2002}.\left(1+2+2^2\right)\right]\)
\(=2.7+2^4.7+....+2^{2002}.7\)
\(=7.\left(2+2^4+....+2^{2002}\right)\)
\(\Rightarrow A⋮7\)