2^1*2^2*2^3*...*2^11=2^(1+2+3+...+11) In  paranteza   ai   o   progresie   geometrica   cu   ratia 2 Calculezi  suma   cu   formula  S11=a1*(2^n-1).(2-r)   unde   n=11   si   a1=1 S11=1*(2^11-1)/(2-1)=(2^11-1) Deci   rezultatul  este 2^(2^11-1)

[latex]2 ^ig1 imes 2 ^ig2 imes2^ig3 imes... imes 2^ig{11} = \ \ = 2^ig{1+2+3+...+11} =\ \ = 2^{dfrac{11cdot(11+1)}{2}} =\ \ = 2^{dfrac{11cdot 12}{2}} = \ \ = 2^ig{11cdot 6} =\ \ =2^ig{66} \ \ \ $M-am folosit de suma lui Gauss ightarrow oxed{1+2+3+...+n = dfrac{ncdot (n+1)}{2}}[/latex]