Series of $\frac{1}{2}(e^{3x} - e^{2x} - e^x + 1)$ I am reading slides from my lecture and saw this equality:
$$  \frac{1}{2}(e^{3x} - e^{2x} - e^x +   \color{red}{1}) = \sum_{r \ge 0}\frac{1}{2}(3^r-2^r-1) \frac{x^r}{r!} $$
but in my opinion it should be
$$  \frac{1}{2}(e^{3x} - e^{2x} - e^x + 1) = \sum_{r \ge 0}\frac{1}{2}(3^r-2^r-1) \frac{x^r}{r!} + \color{red}{\frac{1}{2}} $$
where $  \color{red}{1}$ has been lost?

You are correct, it should be
$$\frac{1}{2}(e^{3x} - e^{2x} - e^x + 1) = \sum_{r \ge 0}\frac{1}{2}(3^r-2^r-1) \frac{x^r}{r!} + \frac{1}{2}.$$
Note that, since the left-hand side is zero for $x=0$, you may also write
$$\frac{1}{2}(e^{3x} - e^{2x} - e^x +   1) = \sum_{r \ge \color{red}{1}}\frac{1}{2}(3^r-2^r-1) \frac{x^r}{r!}$$
where starting index in the sum is now $1$.
