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$x^{n-1} + 2x^{n-2} + 3x^{n-3} + 4x^{n-4} + \cdots + nx^{n-n}$

I'm not sure if this can even be summed. Any help is appreciated.

Martin Sleziak
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jsmith14
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  • How did this come up? – fritzenbauer Nov 22 '15 at 01:23
  • @fritzenbauer I'm trying to find a closed form of a recurrence relation – jsmith14 Nov 22 '15 at 01:24
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    Hint: Do $x=1$ separately. Now for $x\ne 1$ let $f(x)$ be our function. Write down $xf(x)$, and then $xf(x)-f(x)$. Apart from one term, this difference will be a nice geometric series. – André Nicolas Nov 22 '15 at 01:28
  • A few related posts: http://math.stackexchange.com/questions/87030/proof-by-induction-sum-limits-i-0n-i-2i-1-n1-2n-1 http://math.stackexchange.com/questions/11464/how-to-compute-the-formula-sum-r-1d-r-cdot-2r http://math.stackexchange.com/questions/90637/what-is-the-limit-of-sum-limits-n-1-inftyn2-3n http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn – Martin Sleziak Nov 22 '15 at 06:51

3 Answers3

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\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.

Michael Hardy
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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} $.

marty cohen
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  • ok so I know that $\sum_{k=1}^n x^{n-k}$ can be written as $\frac{x^{n}-1}{x-1}$ Is there anyway we could do something similar here? – jsmith14 Nov 22 '15 at 01:34
1

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 $S$.

Gaurang Tandon
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callculus42
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