What is the sum of the series $x^{n-1} + 2x^{n-2} + 3x^{n-3} + 4x^{n-4} + \cdots + nx^{n-n}$? $x^{n-1} + 2x^{n-2} + 3x^{n-3} + 4x^{n-4} + \cdots + nx^{n-n}$
I'm not sure if this can even be summed. Any help is appreciated. 
 A: \begin{align}
& x^{n-1} + 2x^{n-2} + 3x^{n-3} + 4x^{n-4} + \cdots + nx^{n-n} \\[10pt]
& x^{n-1}\left( 1 + \frac 2 x + \frac 3 {x^2} + \frac 4 {x^3} + \cdots + \frac n {x^{n-1}} \right) \\[10pt]
= {} & x^{n-1}\left( 1 + 2y + 3y^2 + 4y^3 + \cdots + ny^{n-1} \right) & (\text{where }y = 1/x) \\[10pt]
= {} & x^{n-1} \frac d {dy} \left( 1 + y + y^2 + y^3 + y^4 + \cdots + y^n \right) \\[10pt]
= {} & x^{n-1} \frac d {dy} \, \frac{1 - y^{n+1}}{1-y}.
\end{align}
Now do the differentiation and then finally put $1/x$ wherever $y$ appears.
A: Hint:
This is
$\begin{array}\\
\sum_{k=1}^n kx^{n-k}
&=x^n\sum_{k=1}^n kx^{-k}\\
\end{array}
$
This sum can be written
in terms of
$\sum_{k=1}^n x^{-k}
$.
A: It is
$S=x^{n-1}+2x^{n-2}+3x^{n-3}+\ldots+(n-2)\cdot x^2+(n-1)\cdot x^1+n\cdot x^0 \quad (1)$
$x\cdot S=x^{n}+2x^{n-1}+3x^{n-2}+\ldots+(n-2)\cdot x^3+(n-1)\cdot x^2+n\cdot x^1 \quad (2)$
Subtracting (1) from (2):
$(x-1)S=\color{blue}{x^n+x^{n-1}+x^{n-2}+x^{n-3}+\ldots + x^2+x}-n$
The formula for the blue expression is $x\cdot \frac{1-x^{n}}{1-x}$ (geometric series)
Now solve for $S$.
