= Solution
Interpret the printed missing parenthesis as ending $f(y_{n+2})$ before the next summand. The <characteristic polynomials of a linear multistep method> are then
$$
\rho(\zeta)=\zeta^3-(1+2\alpha)\zeta^2+(1+2\alpha)\zeta-1
=(\zeta-1)(\zeta^2-2\alpha\zeta+1),
$$
$$
\sigma(\zeta)=\frac\zeta6[(5+\alpha)\zeta^2-(4+8\alpha)\zeta+11-5\alpha].
$$
They satisfy $\rho(1)=0$ and $\rho'(1)=\sigma(1)=2(1-\alpha)$, so the consistency identities hold. The two nonprincipal roots are reciprocal. If $|\alpha|<1$, they are $\alpha\pm i\sqrt{1-\alpha^2}$, distinct unit roots also distinct from the principal root one. If $|\alpha|>1$, one reciprocal root lies outside the unit circle. At $\alpha=1$ the root one is triple, and at $\alpha=-1$ the root minus one is double. The <root condition for a multistep method> therefore holds exactly when $-1<\alpha<1$. The <Dahlquist equivalence theorem>, with standard smoothness and consistent starting data, gives
$$
\boxed{\text{Convergent exactly for }-1<\alpha<1.}
$$
To determine order rather than guess it from the number of steps, the <exponential-symbol order criterion for a multistep method> gives
$$
\rho(e^z)-z\sigma(e^z)=-\frac{\alpha+5}{12}z^4-\frac{14\alpha+61}{90}z^5+O(z^6).
$$
Thus \b[the formal order is three for $\alpha\ne-5$, and four for $\alpha=-5$]. At the exceptional value the $z^5$ coefficient is $1/10$. That fourth-order formula is not zero-stable. Every convergent member consequently has global order three. At $\alpha=1$ the third-order Taylor cancellation is only a formal defect property of a degenerate, nonconvergent recurrence.
Here is a direct <A-stability> test for the convergent parameters. For a nonprincipal unit root $r=\alpha+i\sqrt{1-\alpha^2}$, the simple root of the <amplification polynomial of a multistep method> satisfies
$$
r(z)=r+z\frac{\sigma(r)}{\rho'(r)}+O(z^2).
$$
Reduction using $r^2-2\alpha r+1=0$ gives
$$
\frac{\sigma(r)}{r\rho'(r)}=
\frac{2\alpha^2r+4\alpha r-\alpha-3r-2}{6[(2\alpha+1)r-1]},\qquad
\operatorname{Re}\frac{\sigma(r)}{r\rho'(r)}=\frac{\alpha-1}{12}<0.
$$
For small negative real $z$, it follows that
$$
|r(z)|^2=1+\frac{\alpha-1}{6}z+O(z^2)>1.
$$
So even a small negative real test parameter has an amplifying parasitic mode. \b[None of the convergent parameters is A-stable.] Parameters outside this range already fail the necessary zero-step root condition, so \b[there is no real $\alpha$ giving an A-stable original method].
The endpoint $\alpha=1$ is worth keeping separate from a misleading cancellation: there $\rho=(\zeta-1)^3$ and $\sigma=\zeta(\zeta-1)^2$. The original amplification polynomial retains the double root one for every $z$, giving unbounded polynomial parasitic modes for general starting values. Canceling $(\zeta-1)^2$ would yield a different, reduced <Backward Euler method>; it does not stabilize the original recurrence. This is the <common-factor cancellation defect in a multistep recurrence>. The family is summarized by the <reciprocal-root three-step multistep family>.
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