Consider the binary triple $(0,0,0)$. Now applying a binary addition $(0,0,0)+(1,0,0)=(1,0,0)$. Now apply again to get $(0,0,0)+(1,0,0)+(1,0,0)=(1,0,0)+(1,0,0)=(0,1,0)$.
Repeating this again gives $(0,0,0)+(1,0,0)+(1,0,0)+(1,0,0)=(0,1,0)+(1,0,0)=(1,1,0)$
Repeating this again finally we can land up at the triple (1,1,1). For this case when the size is three I needed 7 steps to arrive at the triple. Is there any general result for no of steps required for $n$-tuple $(0,\cdots,0)$ to arrive at $(1,\cdots,1)$ by adding repeatedly with $(1,0,\cdots,0)$? I am sorry I couldnt write my doubt properly. I tried to prove it by induction. For the $n=4$, the number I am getting is $15=7+7+1$. I verified one more higher case. for $n=5$, it is $31=15+15+1$ related to the $n=4$ case.