Question

fie A=3la puterea 1+3la puterea2+.....+3la puterea2016
ultima cifra a numarului A ESTE?

Answer 1

[tex]A=3^{1}+3^{2}+...+3^{2016}(inmultim\ cu\ 3)\\3*A=3*(3^{1}+3^{2}+...+3^{2016} \\3A=3*3^{1}+3*3^{2}+...+3*3^{2016}\\3A=3^{1+1}+3^{1+2}+...+3^{1+2016}\\3A=3^{2}+3^{3}+...+3^{2017}\\ 3A-A=(3^{2}+3^{3}+...+3^{2017})-(3^{1}+3^{2}+...+3^{2016})\\ 2A=3^{2017}-3\\A=\frac{3^{2017}-3}{2}\\Ultima\ cifra\ a\ puterilor\ lui\ 3\ sunt\ seturi\ de\ 4:3;9;7;1\\2017=4*504+1\\Ultima\ cifra\ a\ lui\ 3^{2017}\ este\ 3;\\3-3=0\\0:2=0\\Ultima\ cifra\ a\ lui\ A=0[/tex]