Finding the variance of a differenced AR(1) process I'm working on an unassessed course problem,

If $y_t$ is the $\text{AR}(1)$ process $$y_t=\alpha_1y_{t-1}+\epsilon_t,$$ where $(\epsilon_t)$ is $\text{White Noise}(0,\sigma^2)$, and $|\alpha_1|<1$, find the covariance function of the differenced process $w_t=y_t-y_{t-1}$, and show that $\text{Var}(w_t)=2\sigma^2/(1+\alpha_1)$. Hence find the acf of $w_t$.

I think I need the variance first, and that will give me the covariance, and that will give me the acf. For the variance I get
\begin{align}
& \begin{aligned}
w_t
& = y_t-y_{t-1} && (1.1) \\
& = (\alpha_1y_{t-1}+\epsilon_t)-y_{t-1} && (1.2) \\
& = (\alpha_1-1)y_{t-1}+\epsilon_t && (1.3) \\
& = \dots \\
& = (\alpha_1-1)\alpha_1^{t-1}y_0+(\alpha_1-1)\sum_{i=0}^{t-2}\alpha_1^i\epsilon_{t-i} && (1.4) \\
& \xrightarrow[|\alpha_1|<1]{t\rightarrow\infty}(\alpha_1-1)\sum_{i=0}^\infty\alpha_1^i\epsilon_{t-i}  && (1.5)
\end{aligned} \\[1em]
\therefore \; & \begin{aligned}[t]
\mathbb{E}(w_t)
& = \mathbb{E}\left((\alpha_1-1)\sum_{i=0}^\infty\alpha_1^i\epsilon_{t-i}\right) && (2.1) \\
& = (\alpha_1-1)\sum_{i=0}^\infty\alpha_1^i\mathbb{E}(\epsilon_{t-i}) && (2.2) \\
& = 0 && (2.3)
\end{aligned} \\[1em]
\therefore \; & \begin{aligned}[t]
\mathbb{V}(w_t)
& = \mathbb{E}(w_t^2)-\mathbb{E}(w_t)^2 && (3.1) \\
& = \mathbb{E}\left(\left((\alpha_1-1)\sum_{i=0}^\infty\alpha_1^i\epsilon_{t-i}\right)^2\right) && (3.2) \\
& = (\alpha_1-1)^2\sum_{i=0}^\infty\alpha_1^{2i}\mathbb{V}(\epsilon_{t-i}) && (3.3) \\
& = (\alpha_1-1)^2\frac{\sigma^2}{1-\alpha_1^2} && (3.4)
\end{aligned}
\end{align}
and I can't figure out how this can equal $2\sigma^2/(1+\alpha_1)$. Could someone give me a pointer?
 A: $\DeclareMathOperator{\Var}{Var}$
$\DeclareMathOperator{\Cov}{Cov}$
In this answer, you have already derived that the auto covariance function of $\{y_t\}$ is
\begin{align}
\gamma(h) = \frac{\sigma^2\alpha_1^h}{1 - \alpha_1^2}.
\end{align}
It then follows from the bilinearity of covariance (see here and here) that
\begin{align}
\Var(w_t) &= \Var(y_t - y_{t - 1}) \\
&= \Var(y_t) + \Var(y_{t - 1}) - 2\Cov(y_t, y_{t - 1}) \\
&= 2\gamma(0) - 2\gamma(1) \\
&= 2\sigma^2\frac{1 - \alpha_1}{1 - \alpha_1^2} \\
&= \frac{2\sigma^2}{1 + \alpha_1}.
\end{align}
A: An AR(1) has covariance
$\gamma(h) = C[y_t, y_{t-h}] = \alpha \gamma(h-1) = \alpha^h \gamma(0)$ and $\gamma(0)
= V[y_t] = \alpha^2\gamma(0) + \sigma^2$ so $\gamma(h) = \alpha^h\frac{\sigma^2}{1-\alpha^2}$.
$$
C[y_t - y_{t-1},y_{t-h} - y_{t-h-1}]
= \gamma(h) - \gamma(h+1) - \gamma(h-1) + \gamma(h)\\
= 2\gamma(h) - \gamma(h+1) - \gamma(h-1)
$$
so
$C[w_t, w_{t-h}] = (2\alpha^h-\alpha^{h+1} -\alpha^{\lvert h-1\rvert })\frac{\sigma^2}{1-\alpha^2}
$
and
$C[w_t, w_{t}] = 2\frac{\sigma^2}{1+\alpha}$.
