\(A=2+2^2+2^3+...+2^{10}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)

\(=3\left(2+2^3+...+2^9\right)⋮3\)

\(A=2+2^2+2^3+...+2^{10}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)

\(=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)\)

\(=\left(2+2^6\right).31⋮31\)

Đặt  A  =\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)    

Ta có \(3A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)

           \(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)

     => \(2A=3A-A=3-\frac{1}{3^{2005}}\)

   => \(A-\frac{3-\frac{1}{3^{2005}}}{2}\)

\(A=2^1+2^2+2^3+...+2^{2010}\)

\(=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{2009}\right)⋮3\)

\(A=2^1+2^2+2^3+...+2^{2010}\)

\(=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+...+2^{2008}\right)⋮7\)

A=21+22+23+...+22010

=(21+22)+(23+24)+...+(22009+22010)=(21+22)+(23+24)+...+(22009+22010)

=2(1+2)+23(1+2)+...+22009(1+2)=2(1+2)+23(1+2)+...+22009(1+2)

=3(2+23+...+22009)⋮3=3(2+23+...+22009)⋮3

�=21+22+23+...+22010A=21+22+23+...+22010

=(21+22+23)+(24+25+26)+...+(22008+22009+22010)=(21+22+23)+(24+25+26)+...+(22008+22009+22010)

=2(1+2+22)+24(1+2+22)+...+22008(1+2+22)=2(1+2+22)+24(1+2+22)+...+22008(1+2+22)

=7(2+24+...+22008)⋮7=7(2+24+...+22008)⋮7

sao ko dung f(x) ma viet

\(a=2+2^2+2^3+2^4+2^5+2^6+2^7+2^9+2^{10}\)

a=\(\left(2+2^2\right)+2^2.\left(2+2^2\right)+..+2^8\left(2+2^2\right)\)

a=\(\left(2+2^2\right).\left(1+2^2+..+2^8\right)\)

a=\(6.\left(1+2^2+2^4+2^6+2^8\right)\)

chia het cho 3

Bằng 365:365=1

Là vậy đó bạn