Question

Valoare nr . 2 ^1 x 2 ^2 ............ 2^11

Answer 1

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)

Answer 2

$2 ^\big1 \times 2 ^\big2\times2^\big3 \times...\times 2^\big{11} = \\ \\ = 2^\big{1+2+3+...+11} =\\ \\ = 2^{\dfrac{11\cdot(11+1)}{2}} =\\ \\ = 2^{\dfrac{11\cdot 12}{2}} = \\ \\ = 2^\big{11\cdot 6} =\\ \\ =2^\big{66} \\ \\ \\ M-am folosit de suma lui Gauss \rightarrow \boxed{1+2+3+...+n = \dfrac{n\cdot (n+1)}{2}}$