= Solution
For a decaying <bound state> take $\alpha>0$. Substituting $F=fe^{-\alpha r}$ gives $F''=e^{-\alpha r}(f''-2\alpha f'+\alpha^2f)$. The term involving the <energy> cancels $\alpha^2f$, leaving the displayed equation with
$$
\boxed{A=\frac{2mK}{\hbar^2}.}
$$
At the lowest power of a series from the <Frobenius method> $f=\sum_{k\geq0}a_kr^{k+c}$, the <indicial equation> is $c(c-1)-\ell(\ell+1)=0$. Its roots are $\ell+1$ and $-\ell$. The latter makes $F\sim r^{-\ell}$ and the <radial probability density> integral $\int|F|^2dr$ diverge for positive integral $\ell$. Hence $\boxed{c=\ell+1}$.
Equating the remaining coefficients gives
$$
\boxed{a_{k+1}=\frac{2\alpha(k+\ell+1)-A}{(k+1)(k+2\ell+2)}a_k.}
$$
Unless the numerator vanishes, the regular confluent-hypergeometric series contains a large-radius growing branch with the exponential factor $e^{2\alpha r}$. Multiplication by the external $e^{-\alpha r}$ then leaves exponential growth, so it cannot be a <normalizable> <bound state>. Equivalently, the recurrence at large $k$ has $a_{k+1}/a_k\sim2\alpha/k$, indicating that growing exponential scale; regularity at the origin has already fixed the combination of solutions, leaving termination as the way to remove it.
If the polynomial ends at degree index $N\geq0$, its last recurrence numerator gives $A=2\alpha(N+\ell+1)$. With $n=N+\ell+1>\ell$,
$$
\boxed{\alpha=\frac A{2n},\qquad E_n=-\frac{\hbar^2A^2}{8mn^2}=-\frac{mK^2}{2\hbar^2n^2}.}
$$
For the separate angle-independent exponential, <normalization> includes the full <solid angle>:
$$
1=4\pi|N|^2\int_0^\infty r^2e^{-2\alpha r}\,dr
=4\pi|N|^2\frac{2}{(2\alpha)^3}=\frac{\pi|N|^2}{\alpha^3}.
$$
Thus one can choose $\boxed{N=\sqrt{\alpha^3/\pi}}$, up to an overall <phase>. Angle independence corresponds to the $\ell=0$ extension, distinct from the positive-$\ell$ family used in the indicial discussion.
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