A regular $ n\minus{}$gon $ A_1A_2A_3 \cdots A_k \cdots A_n$ inscribed in a circle of radius $ R$ is given. If $ S$ is a point on the circle, calculate \[ T \equal{} \sum^n_{k\equal{}1} SA^2_k.\]
1. **Setup the problem in the complex plane:**
   - Assume the regular $n$-gon $A_1A_2A_3 \cdots A_k \cdots A_n$ is inscribed in a circle of radius $R$.
   - Let the center of the circle be the origin of the complex plane.
   - Let $a_k$ be the complex number representing the vertex $A_k$, and let $s$ be the complex number representing the point $S$ on the circle.

2. **Express the vertices and point $S$ in complex form:**
   - The vertices $A_k$ can be represented as $a_k = R e^{i \frac{2k\pi}{n}}$ for $k = 1, 2, \ldots, n$.
   - The point $S$ can be represented as $s = R e^{i \theta}$ for some angle $\theta$.

3. **Calculate the squared distance $|s - a_k|^2$:**
   \[
   |s - a_k|^2 = |R e^{i \theta} - R e^{i \frac{2k\pi}{n}}|^2 = R^2 |e^{i \theta} - e^{i \frac{2k\pi}{n}}|^2
   \]

4. **Simplify the expression inside the sum:**
   \[
   |e^{i \theta} - e^{i \frac{2k\pi}{n}}|^2 = (e^{i \theta} - e^{i \frac{2k\pi}{n}})(\overline{e^{i \theta}} - \overline{e^{i \frac{2k\pi}{n}}})
   \]
   Since $\overline{e^{i \theta}} = e^{-i \theta}$ and $\overline{e^{i \frac{2k\pi}{n}}} = e^{-i \frac{2k\pi}{n}}$, we have:
   \[
   |e^{i \theta} - e^{i \frac{2k\pi}{n}}|^2 = (e^{i \theta} - e^{i \frac{2k\pi}{n}})(e^{-i \theta} - e^{-i \frac{2k\pi}{n}})
   \]

5. **Expand and simplify the product:**
   \[
   (e^{i \theta} - e^{i \frac{2k\pi}{n}})(e^{-i \theta} - e^{-i \frac{2k\pi}{n}}) = 1 - e^{i (\theta - \frac{2k\pi}{n})} - e^{-i (\theta - \frac{2k\pi}{n})} + 1
   \]
   Using Euler's formula, $e^{i x} + e^{-i x} = 2 \cos x$, we get:
   \[
   1 - e^{i (\theta - \frac{2k\pi}{n})} - e^{-i (\theta - \frac{2k\pi}{n})} + 1 = 2 - 2 \cos (\theta - \frac{2k\pi}{n})
   \]
   Therefore,
   \[
   |e^{i \theta} - e^{i \frac{2k\pi}{n}}|^2 = 2 - 2 \cos (\theta - \frac{2k\pi}{n})
   \]

6. **Sum the squared distances:**
   \[
   T = \sum_{k=1}^n |s - a_k|^2 = R^2 \sum_{k=1}^n |e^{i \theta} - e^{i \frac{2k\pi}{n}}|^2 = R^2 \sum_{k=1}^n (2 - 2 \cos (\theta - \frac{2k\pi}{n}))
   \]

7. **Evaluate the sum:**
   \[
   T = R^2 \sum_{k=1}^n (2 - 2 \cos (\theta - \frac{2k\pi}{n})) = 2R^2 \sum_{k=1}^n (1 - \cos (\theta - \frac{2k\pi}{n}))
   \]
   The sum of cosines over a full period is zero:
   \[
   \sum_{k=1}^n \cos (\theta - \frac{2k\pi}{n}) = 0
   \]
   Therefore,
   \[
   T = 2R^2 \sum_{k=1}^n 1 = 2R^2 \cdot n = 2nR^2
   \]

The final answer is $\boxed{2nR^2}$.