Ta có : \(X=1+2^2+2^4+.....+2^{2010}\)
\(\Rightarrow2^2X=2^2+2^6+2^8+.....+2^{2012}\)
\(4X=2^2+2^6+2^8+.....+2^{2012}\)
\(4X-X=2^{2012}-1\)
\(3X=2^{2012}-1\)
\(X=\frac{2^{2012}-1}{3}\) (sai đề nhé )
ta có: X=\(1+2+2^2...2^{2010}\Rightarrow2X=2+2^2+...2^{2011}\)
\(\Rightarrow2X-X=\left(2+2^2...2^{2011}\right)-\left(1+2+...2^{2010}\right)\)
\(\Rightarrow X=2^{2011}-1\)
xét hiêu Y-X=\(2^{2011}-\left(2^{2011}-1\right)=1\)
vậy X,Y là 2 số tự nhiên liên tiếp