Question

Please, solve this exercice:

1+2+2²+2³+2⁴+....+ 2²⁰¹¹

It can be solved by using geometric progression?

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=(a_1(1-r^n))/(1-r)\\\\a_1=1;\ r=2\\\\subtitute:\\\\S_(2012)=(1(1-2^(2012)))/(1-2)=(1-2^(2012))/(-1)=2^(2012)-1\\\\Only\ that...(606\ digits,\ if\ you\ want\ how\ length\ this\ number)`