Question:

Step 1: The sum is given by $$S = 1^2 + 3\alpha + 5\alpha^2 + 7\alpha^3 + \cdots + (2n-1)\alpha^{n-1}$$ This is an arithmetic-geometric series. Step 2: Multiply $S$ by $\alpha$: $$\alpha S = \alpha + 3\alpha^2 + 5\alpha^3 + \cdots + (2n-3)\alpha^{n-1} + (2n-1)\alpha^n$$ Step 3: Subtract $\alpha S$ from $S$: $$S - \alpha S = 1 + (3-1)\alpha + (5-3)\alpha^2 + \cdots + ((2n-1)-(2n-3))\alpha^{n-1} - (2n-1)\alpha^n$$ $$S(1 - \alpha) = 1 + 2\alpha + 2\alpha^2 + \cdots + 2\alpha^{n-1} - (2n-1)\alpha^n$$ $$S(1 - \alpha) = 1 + 2(\alpha + \alpha^2 + \cdots + \alpha^{n-1}) - (2n-1)\alpha^n$$ Step 4: Since $\alpha$ is an $n$-th root of unity and $\alpha \neq 1$, the following properties hold: $$\alpha^n = 1$$ $$1 + \alpha + \alpha^2 + \cdots + \alpha^{n-1} = 0$$ From the second property, it follows that: $$\alpha + \alpha^2 + \cdots + \alpha^{n-1} = -1$$ Step 5: Substitute these properties into the expression for $S(1 - \alpha)$: $$S(1 - \alpha) = 1 + 2(-1) - (2n-1)(1) + (2n-1)(1)$$ $$S(1 - \alpha) = 1 - 2 - (2n-1) + (2n-1)$$ $$S(1 - \alpha) = 1 - 2$$ $$S(1 - \alpha) = -1$$ Step 6: Solve for $S$: $$S = -\frac{1}{1 - \alpha}$$