The busy beaver function is well known, but to be specific, let me define the busy beaver function BB(n)= maximum number of 1's that can be printed on the tape by a standard n-state, 2-symbol halting Turing machine before halt.

My question maybe trivial: is it true that BB(n+1)>BB(n) for all n?

The claim is true if we measure the busy beaver by BB'(n)= maximum number ofshiftsmade by the machine before halt. (This is because we can always change the last shift made by BB'(n), say $\displaystyle \{ 0{q}_{i}1RH\}$, to $\displaystyle \{ 0{q}_{i}1R{q}_{n+1}\}$ and $\displaystyle \{1{q}_{n+1}1RH\}$, thus making sure that BB'(n+1)$\displaystyle \geq$BB'(n)+1$\displaystyle >$BB'(n)). Can we show that the strict inequality holds too if we measure byscoresinstead of by shifts?