Question

Se dau numerele a=2+2·3+2·3²+2·3³+...+3^2016 si b=8·3^2017+1.Aratati ca a+b este cub perfect

Answer 1

Rezoovarea in atasament

Answer 2

   
[tex]\displaystyle\\ a=2\cdot3^0+2\cdot3^1+2\cdot3^2+2\cdot3^3+...+2\cdot3^{2016}=\\\\ =2(3^0+3^1+3^2+3^3+...+3^{2016}) =\\\\ =2\times\frac{3^{2016+1}-1}{3-1}=2\times\frac{3^{2017}-1}{2} =\boxed{3^{2017}-1}\\\\ b=\boxed{8\cdot3^{2017}+1}\\\\ a+b=3^{2017}-\underline{1}+8\cdot3^{2017}+\underline{1}=3^{2017}+ 8\cdot3^{2017}=3^{2017}(1+8)=\\\\ =9\cdot3^{2017}=3^2\cdot3^{2017}=3^{2+2017}=3^{2019}=3^{673\times3}=\boxed{\Big(3^{673}\Big)^3 =cp}[/tex]