# 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?

• Are $x$ and $r$ the same variable? – José Carlos Santos Aug 11 '19 at 11:23

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 $$1$$.
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.
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.