[tex]\boxed{\boxed{1\cdot 2^0+2\cdot2^1+3\cdot2^2+4\cdot2^3+...+n\cdot2^{n-1} = 1+(n-1)\cdot 2^n}} \\ \\ \\ \Rightarrow 1+2\cdot2^1+3\cdot2^2+4\cdot2^3+...+2016\cdot2^{2015} = \\ =1\cdot 2^0+2\cdot2^1+3\cdot2^2+4\cdot2^3$+...+$2016\cdot2^{2016-1}$= 1+(2016-1)\cdot 2^{2016}= \\ =1+2015\cdot 2^{2016} \\ \\ \Rightarrow \boxed{1+2\cdot2^1+3\cdot2^2+4\cdot2^3+...+2016\cdot2^{2015} = 1+2015\cdot 2^{2016}}[/tex]
Nu stiu cum sa demonstrez acea formula... Poate asta voiai sa afli de fapt..
