first of all, if i were the OP, i would fix the notation a bit. $x(t)$, $y(t)$ are continuous-time signals and $t\in\mathbb{R}$ is a continuous argument. $x[n] \triangleq x(nT)$ and $y[n] \triangleq y(nT)$ are discrete=time signals and and $n\in\mathbb{Z}$ is a discrete argument, an integer. square brackets mean that an integer goes into them. the normal convention is that $T=\frac{1}{f_s}$ is your sampling period.
a simple moving-average filter with equal-weighting in the average is
$$ \begin{align}
y[n] & = \frac{1}{N} \sum\limits_{i=0}^{N-1} x[n-i] \\
& = \frac{1}{N} \left(x[0] + \sum\limits_{i=1}^{N} x[n-i] - x[n-N]\right) \\
& = \frac{1}{N} \sum\limits_{i=1}^{N} x[n-i] + \frac{1}{N} \left(x[0] - x[n-N]\right) \\
& = \frac{1}{N} \sum\limits_{i=0}^{N-1} x[n-1-i] + \frac{1}{N} \left(x[0] - x[n-N]\right) \\
& = y[n-1] + \frac{1}{N} \left(x[0] - x[n-N]\right) \\
\end{align} $$
that is representing both the non-recursive implementation and the recursive. the recursive implementation works as long as you subtract exactly what you earlier added, something that might not happen in a floating-point context. both are FIR, even though the recursive implementation is something we would call a Truncated IIR, still an FIR.
to get the frequency response, apply the Z-Transform to either form, and solve for $Y(z)$ in terms of $X(z)$. then compute the transfer function:
$$ H(z) \triangleq \frac{Y(z)}{X(z)} $$
then substitute $z = e^{j \omega}$ and solve for the $\omega$ such that $\left|H(e^{j \omega})\right|^2 = \frac{1}{2}$. that $\omega= 2\pi f T$ is your -3.01 dB cut-off frequency.
try doing that math yourself. we'll be here to help guide you.
