(2013^0+1+2+2²+2³+....+2^100)·x=
=(1+1+2+2²+2³+....+2^100)·x=
=(2+2+2²+2³+....+2^100)·x= ( 2+2+2²=2²+2²=2·2²=2³)
= (2³+2³+....+2^100)·x=
=(2^4+....+2^100)·x=
................................
=(2^100+2^100)·x
=2^101·x
Acum putem scrie
2^101·x=2^101
x=1
