I’m struggling with how to express the solution to the question in terms of the variance.
this is the solution$$\mathbb{E}(Y_n Y_{n+m}) = \sigma_V^2 \cdot [(1+n(n+m))\delta(m) +n\delta(m+1)+ (n+m)\delta(m-1)]$$ whereas I got $$R_{Y}(n,n+m) = R_V(m) (n^2+1) + nR_V(m-1) + nR_V(m+1)$$
Since $V_n$ are i.i.d., note that $Y_n$ and $Y_{n +m}$ are independent for $|m| \geq 2$. Consequently, $\mathbb{E}[Y_n Y_{n+m}] = \mathbb{E}[Y_n] \mathbb{E}[Y_{n+m}] = 0$ as $\mathbb{E}[Y_n] = 0$ for all $n$.
For $m = 0$, we have, \begin{align*} \mathbb{E}[Y_n^2] = \mathbb{E}[(V_n + a V_{n-1})^2] = \mathbb{E}[V_n^2] + a^2 \mathbb{E}[V_{n-1}^2] + 2a \mathbb{E}[V_n] \mathbb{E}[V_{n-1}] = \sigma_V^2 (1 + a^2). \end{align*}
For $m = 1$, we have, \begin{align*} \mathbb{E}[Y_n Y_{n+1}] & = \mathbb{E}[(V_n + a V_{n-1})(V_{n+1} + a V_{n})] \\ & = \mathbb{E}[V_n] \mathbb{E}[V_{n+1}] + a \mathbb{E}[V_n^2] + a \mathbb{E}[V_{n+1}] \mathbb{E}[V_{n-1}] + a^2 \mathbb{E}[V_n] \mathbb{E}[V_{n-1}] \\ & = a \sigma_V^2. \end{align*}
The $m = -1$ is identical to that of $m = 1$. On combining all of them, we obtain, \begin{align*} \mathbb{E}[Y_n Y_{n+m}] = \sigma_V^2 [a^2 \delta(m) + a\delta(m+1) + a\delta(m-1)]. \end{align*}
I am not sure why the solution shows a dependence on $n$ and $m$, apart from the $\delta$ functions.