1. **State the problem:** Find the value of the sum $$\sum_{r=1}^n {^nC_r} \alpha_r$$ where $$\alpha_r = e^{i 2 (r-1) \pi / n}$$ are the $n$th roots of unity. 2. **Recall the definitions:** The $n$th roots of unity satisfy: $$\alpha_r = e^{i 2 (r-1) \pi / n}, \quad r=1, 2, \ldots, n.$$ Also, the binomial coefficient $^nC_r$ is the number of ways to choose $r$ elements from $n$. 3. **Look at the sum:** $$S = \sum_{r=1}^n {^nC_r} \alpha_r.$$ We want to express this sum in terms of $\alpha_1$ and other known quantities. 4. **Consider the binomial expansion:** Recall the binomial theorem: $$ (1 + x)^n = \sum_{r=0}^n {^nC_r} x^r. $$ If we let $x = \alpha$, then: $$ (1 + \alpha)^n = \sum_{r=0}^n {^nC_r} \alpha^r. $$ 5. **Relate $\alpha_r$ to powers of $\alpha = \alpha_2$:** Since $\alpha_r = e^{i 2 (r-1) \pi / n}$, we can write: $$\alpha_r = \alpha^{r-1}$$ where $\alpha = e^{i 2 \pi / n} = \alpha_2$. 6. **Rewrite the sum in powers of $\alpha$: ** $$ S = \sum_{r=1}^n {^nC_r} \alpha^{r-1} = \sum_{r=0}^{n-1} {^nC_{r+1}} \alpha^r. $$ 7. **Express $S$ in terms of $(1 + \alpha)^n$:** Note that: $$ (1 + \alpha)^n = \sum_{r=0}^n {^nC_r} \alpha^r = {^nC_0} + \sum_{r=1}^n {^nC_r} \alpha^r = 1 + \sum_{r=1}^n {^nC_r} \alpha^r. $$ We can rewrite: $$ \sum_{r=1}^n {^nC_r} \alpha^r = (1 + \alpha)^n - 1. $$ 8. **Relate it back to $S$: ** Our sum uses powers $\alpha^{r-1}$, so multiply both sides by $1/\alpha$: $$ S = \sum_{r=1}^n {^nC_r} \alpha^{r-1} = \frac{1}{\alpha} \sum_{r=1}^n {^nC_r} \alpha^r = \frac{(1 + \alpha)^n - 1}{\alpha}. $$ 9. **Since $\alpha = \alpha_2$, and $\alpha_1 = 1$, substitute back:** $$ S = \frac{(1 + \alpha_2)^n - 1}{\alpha_2}. $$ 10. **Compare with the given options:** Option B is: $$ \frac{\alpha_1}{2} [(1 + \alpha_1)^n - 1] $$ Since $\alpha_1 = 1$, this equals: $$ \frac{1}{2} [(1 + 1)^n - 1] = \frac{2^n - 1}{2} $$ which is not the same. Option A is: $$ (1 + \frac{\alpha_2}{\alpha_1})^n - 1 = (1 + \alpha_2)^n - 1 $$ which is close but missing division by $\alpha_2$. None of the options match exactly $\frac{(1 + \alpha_2)^n - 1}{\alpha_2}$. However, noticing the problem probably intends $\alpha_1=1$, option B resembles the formula but with a factor 1/2 and $\alpha_1$ in numerator instead of denominator. Given standard formula for this sum is: $$\sum_{r=1}^n {^nC_r} \alpha_r = \frac{(1+\alpha)^n - 1}{\alpha}$$ with $\alpha = \alpha_2$ and option B can be rewritten using $\alpha_1=1$ as: $$\frac{1}{2}[(1 + 1)^n - 1] = \frac{2^n - 1}{2}$$ which differs. Hence, the correct simplified form matches option B in the problem statement only if $\alpha_1$ is 1 and there may be a typo there. **Final answer:** $$\sum_{r=1}^n {^nC_r} \alpha_r = \frac{(1 + \alpha_2)^n - 1}{\alpha_2}.$$