Calculate $\sum_{i=1}^{n-1} i\alpha^{2i}$

Question

I try to calculate $\sum_{i=1}^{n-1} i\alpha^{2i}$ I think one week but still have no ideas to this sigma,thank everyone

Answer 1

You wish to calculate $$ \sum_{i = 1}^{n-1} i (\alpha^2)^i $$ which is $$ \alpha^2 \sum_{i=1}^{n-1}i(\alpha^2)^{i-1} $$ which is the value at $x = \alpha^2$ of $$ \alpha^2 \frac{d}{dx} \sum_{i = 1}^{n-1} x^i $$ which is the value at $x = \alpha^2$ of $$ \alpha^2 \frac{d}{dx} \frac{x^n - 1}{x - 1} $$ (we've added a constant term $1$ to the sum, which is ok since we're going to differentiate it). This is the value at $x = \alpha^2$ of $$ \alpha^2 \frac{nx^{n-1}(x-1) - (x^n - 1)}{(x-1)^2} $$ and now plug in $x = \alpha^2$ and simplify.

Answer 2

Call $S = \sum_{i=1}^{n-1} i \alpha^{2i}=\sum_{i=1}^{n-1} i (\alpha^2)^{i}$ and $\alpha^2 = \beta$. If $\beta \neq 1$, then \begin{align} \beta S = \sum_{i=1}^{n-1} i \beta^{i+1} &= \sum_{i=1}^{n-1}\left((i+1)-1\right) \beta^{i+1} \\ &=\sum_{i=1}^{n-1} (i+1)\beta^{i+1} - \sum_{i=1}^{n-1} \beta^{i+1} \\ &= \sum_{i=2}^{n}i \beta^{i} - \sum_{i=2}^{n}\beta^{i} \\ &= \left( n\beta^n-\beta \right) +\sum_{i=1}^{n-1}i\beta^{i} - \sum_{i=2}^{n}\beta^{i} \\ &= \left( n\beta^n-\beta \right) + S - \frac{\beta^{n+1} - \beta^2 }{\beta -1 } \end{align}

Thus $$ (\beta - 1)S = n\beta^{n}-\beta - \frac{\beta^{n+1} - \beta^2}{\beta -1 } $$

or \begin{align} S &= \frac{n\beta^n -\beta}{\beta -1} - \frac{\beta^{n+1} - \beta^2 }{(\beta - 1)^2 } \\ &=\frac{(n-1)\beta^{n+1} - n\beta^n + \beta}{ (\beta - 1)^2} \\& = \frac{(n-1)\alpha^{2n+2} - n\alpha^{2n} + \alpha^2}{ (\alpha^2 - 1)^2} \end{align}

If $\beta = \alpha^2 = 1$, then $S=\sum_{i=1}^{n-1} i (\alpha^2)^{i} = \sum_{i=1}^{n-1}i=\frac{(n-1)n}{2}$.