Here you have 10 terms due to the bearing rate change (from 0.01 to 0.1). However, if you assume stepwise increments, there should be 9 terms with the last one being 0.09. Indeed, in your calculation in the previous quote, for time +1 the rate change did not make a contribution, for time +2 the contribution was 0.01, for time +3 it was 0.01 + 0.02.

If we assume that the bearing rate, which I denoted $\displaystyle \omega$, changes continuously, let's see how it depends on time.

Initially it is $\displaystyle b$ and then it grows linearly with $\displaystyle t$, so at time $\displaystyle t$ it becomes $\displaystyle b+ct$. The actual bearing is the shaded area under the graph (e.g., when $\displaystyle c = 0$, it is just $\displaystyle bt$). The triangle area is $\displaystyle ct^2/2$ (half of a rectangle with sides $\displaystyle t$ and $\displaystyle ct$), so the shaded area is $\displaystyle bt + ct^2/2$. Therefore, the bearing at time $\displaystyle t$ is $\displaystyle a+bt+ct^2/2$.

If you insist on stepwise increment to the bearing rate, then you can use

the formula for the sum of arithmetic progression: $\displaystyle c+2c+\dots+ (t-1)c=c(1+2+\dots+t-1)=c(t-1)t/2$ when $\displaystyle t$ is an integer and $\displaystyle t>0$. Then the bearing at time $\displaystyle t$ is $\displaystyle a+bt+c(t-1)t/2$.