Series of $\frac{1}{2}(e^{3x} - e^{2x} - e^x + 1)$

Question

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?

Answer 1

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 $1$.

Answer 2

The value of the function at $0$ is $0$, so the constant term of the Maclaurin expansion cannot be anything but $0$. Therefore, since computing the first term of the sequence as it was printed gives $-\frac12$, $\frac12$ should be added to compensate, as you suspect.

Answer 3

You are correct the $1$ term was "lost". For each $e^{nx}$ term on the left, there's a corresponding $\sum_{r\ge 0}n^r\left(\frac{x^r}{r!}\right)$ (I assume your "$t$" is meant to be "$x$") term on the right. Thus, the $\frac{1}{2}(e^3x - e^{2x} - e^x)$ is accommodated by the terms on the right, so the $ + 1$ term is, as you state, not included and, thus, should give an extra $\frac{1}{2}$ term.

An easy way to check is to use $x = 0$ on the left. This gives $0$ but, on the right with $x^0 = 1$, you get $\frac{1}{2}(1 - 1 - 1) = -\frac{1}{2}$ instead.