If alpha neq 1 is any n^{th} root of unity,th | vedclass

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(B) Given $S = 1 + 3\alpha + 5\alpha^2 + \dots + (2n - 1)\alpha^{n-1}$ $(i)$Multiply by $\alpha$: $\alpha S = \alpha + 3\alpha^2 + 5\alpha^3 + \dots +

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$-\frac{2n}{1 - \alpha}$

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(B) Given $S = 1 + 3\alpha + 5\alpha^2 + \dots + (2n - 1)\alpha^{n-1}$ $(i)$Multiply by $\alpha$: $\alpha S = \alpha + 3\alpha^2 + 5\alpha^3 + \dots + (2n - 1)\alpha^n$ $(ii)$Subtracting $(ii)$ from $(i)$:$(1 - \alpha)S = 1 + 2\alpha + 2\alpha^2 + \dots + 2\alpha^{n-1} - (2n - 1)\alpha^n$Since $\alpha^n = 1$ for any $n^{th}$ root of unity:$(1 - \alpha)S = 1 + 2(\alpha + \alpha^2 + \dots + \alpha^{n-1}) - (2n - 1)$Using the sum of geometric progression $\alpha + \alpha^2 + \dots + \alpha^{n-1} = \frac{\alpha(1 - \alpha^{n-1})}{1 - \alpha} = \frac{\alpha - 1}{1 - \alpha} = -1$:$(1 - \alpha)S = 1 + 2(-1) - 2n + 1 = 1 - 2 - 2n + 1 = -2n$Therefore,$S = -\frac{2n}{1 - \alpha}$.

If $\alpha \neq 1$ is any $n^{th}$ root of unity,then $S = 1 + 3\alpha + 5\alpha^2 + \dots$ up to $n$ terms,is equal to

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