Obtaining the pmf of a function while knowing its mgf $
\newcommand{\P}{\mathbb{P}}
\newcommand{\E}{\mathbb{E}}
\newcommand{\Z}{\mathbb{Z}}
$The Problem: This problem is given as Exercise 4 in Chapter 3 of An Intermediate Course in Probability, by Allen Gut. Given that $Y$ is an r.v. with
$$
\E\left[Y^k\right] = \frac 1 4 + 2^{k-1}, \forall k \in \Z^+
$$
we seek to find the distribution of $Y$, namely its pmf.

Attempts/Context:
This chapter of the text focuses on the basics of probability and moment generating functions, and characteristic functions. Some Googling suggests that one may be able to recover the pmf from these by using inverse Fourier or Laplace transforms, but those seem a bit too much to be really kosher.
Now, calculating the mgf $\psi_Y$ of $Y$ at least is trivial enough because we know
$$\psi_Y(t) = 1 + \sum_{n \ge 1} \frac{t^n}{n!} \cdot \E \left[ Y^n \right]$$
and, skipping the algebra for simplicity, we see that the mgf of our given r.v. $Y$ is
$$\psi_Y(t) = \frac 1 4 + \frac 1 4 e^t + \frac 1 2 e^{2t}$$
Other Googling suggests that, owing to this pattern, we can conclude that
$$\P(Y=y) = \begin{cases}
1/4 & y = 0, 1 \\
1/2 & y = 2 \\
0 & \text{otherwise} \end{cases}$$
which matches up with the text's answer. In general, if we had, say,
$$\psi_X(t) = \sum \alpha_k e^{kt}$$
with $\sum \alpha_k = 1$, then I guess the pattern would be that $\P(X=k) = \alpha_k$.
However, this pattern also is not mentioned in the text, and I'm not sure how to justify this from a more elementary perspective.
Can anyone give me ideas on how to properly obtain this answer? Maybe there's some easy use of the definition, or a theorem I'm overlooking...
 A: Yep, you're almost finished! The hard part was computing that
$$ \psi_Y(t) = \frac 1 4 + \frac 1 4 e^t + \frac 1 2 e^{2t}; \tag{$\ast$}$$
and remember that the definition of the moment generating function is
$$ \psi_Y(t) = \mathbb{E}(e^{ty}) .$$
Now, since we are finding a pmf then we have a discrete random variable and so we appeal to the discrete definition of $\psi_Y(t)$. That is, $$ \psi_Y(t) = \mathbb{E}(e^{ty}) = \sum_{y=0}^\infty e^{ty}\,\mathbb{P}(Y=y). $$
But this must be consistent with $(\ast)$ and so, $$\sum_{y=0}^\infty e^{ty}\,\mathbb{P}(Y=y) = \frac 1 4 + \frac 1 4 e^t + \frac 1 2 e^{2t}. $$
Suppose that we only compute the first term of our sum, this tells us that
$ e^{0\cdot y}\,\mathbb{P}(Y=0) = \mathbb{P}(Y=0). $
Similarly, we compute the second term of our sum to give us $ e^{1\cdot y}\,\mathbb{P}(Y=1) = e^y\,\mathbb{P}(Y=1). $
Finally, we compute the third term to give us $ e^{2\cdot y}\,\mathbb{P}(Y=2) = e^{2y}\,\mathbb{P}(Y=2). $
All we need to do now is equate coefficients of $e^{ty}$ to give us the required pmf. That is,
$$ \mathbb{P}(Y=0) + e^t\,\mathbb{P}(Y=1) + e^{2t}\,\mathbb{P}(Y=2) + \sum_{y=3}^\infty e^{ty}\,\mathbb{P}(Y=y) = \frac 1 4 + \frac 1 4 e^t + \frac 1 2 e^{2t}.$$
We discard $y\geq 3$ since all coefficients of these terms are $0$.
