jmc

The left side of the equation, $T(b+1)-T(b)$, gives $$\frac{(b+1)(b+2)}{2}-\frac{b(b+1)}{2},$$which simplifies to $$\frac{b^2+3b+2-b^2-b}{2}=\frac{2b+2}{2}=b+1.$$That is, $b+1$ is equal to $T(x)$, a triangular number.

Since $b>2011$, we are looking for the the smallest triangular number greater than 2012.

After some trial and error, we observe that $T(62)=1953$ and $T(63)=2016$, and so $b+1=2016$ or $b=\boxed{2015}$ is the smallest value that works.