Question

Please, solve this exercice:

Answer 1

$1+2+2^2+2^3+2^4+...+2^{2011}\\\\a_1=1;\ a_2=1\cdot2=2;\ a_3=2\cdot2=2^2;\ a_4=2^2\cdot2=2^3\\\vdots\\a_{2012}=2^{2010}\cdot2=2^{2011}$

$The\ sum\ of\ a\ terms\ of\ geometric\ progression:S_n=\frac{a_1(1-r^n)}{1-r}\\\\a_1=1;\ r=2\\\\subtitute:\\\\S_{2012}=\frac{1(1-2^{2012})}{1-2}=\frac{1-2^{2012}}{-1}=2^{2012}-1\\\\Only\ that...(606\ digits,\ if\ you\ want\ how\ length\ this\ number)$