Question

calculati :2*(1+3+3^2+...+3^49)

Answer 1

   
[tex]\displaystyle \\ 2\times(1 + 3 +3^2 + \hdots + 3^{49}) = \\ \\ 2\times(3^0 + 3^1 +3^2 + 3^3 \hdots +3^{48}+ 3^{49}) = 2\times \frac{3^{50}-1}{3-1} = \boxed{3^{50}-1} \\ \\ [/tex]