\(A=2^1+2^2+2^3+2^4+...+2^{2009}+2^{2010}\)(có 2010 số hạng)
\(A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)(có 1005 nhóm)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)\)
\(A=3\cdot\left(2+2^3+...+2^{2009}\right)⋮3\)(1)
\(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)(có 670 nhóm)
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(A=\left(1+2+2^2\right)\left(2+2^4+...+2^{2008}\right)\)
\(A=7\cdot\left(2+2^4+...+2^{2008}\right)⋮7\left(2\right)\)
Từ (1) và (2)\(\Rightarrow A⋮3;7\)
Ta có : A = 21+22+23+24+...+22010
= (2+22) + (23+24) +... + (22009+22010)
= (2.1+2.2) + (23.1+23.2) + ... + (22009.1 = 22009.2)
= 2 . (1+2) + 23 . (1+2) + ... + 22009 . (1+2)
= 2 . 3 + 22 . 3 + ... + 22009 . 3
=> Chia hết cho 3
Và ta lại có : A = 21+22+23+24+... +22009
= (21+22+23) + (24+25+26) + (22008 + 22009 + 22010)
= (2.1+2.2+2.22) + (24.1+24.2+24.22) +...+ (22008.1+22008.2+22008.22)
= 2 . (1+2+4) + 24 . (1+2+4) +...+22008 . (1+2+4)
= 2 . 7 + 24 . 7 + 22008 . 7
=> Chia hết cho 7
Vay A = 21 + 22 + 23 + 24+ ... + 22010 chia hết cho 3 ; cho 7.
A = 21 + 22 + 23 + 24 +...+ 22010
= (21 + 22 + 23) + (24 + 25 + 26) + ... + (22008 + 22009 + 22010)
= 21(1 + 2 + 22) + 24(1 + 2 + 22) + ... + 22008(1 + 2 + 22)
= 21.7 + 24.7 +... + 22008.7
= 7(21 + 24 + ... + 22008)
Vì 7(21 + 24 + ... + 22008) ⋮ 7 nên A ⋮ 7
