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How can I solve this sum?

$$S_{n} = \sum_{k=1}^{n} k \Big( \frac{3}{2} \Big)^{k-1}$$

Worth noting: I can use lower index $k = 0$ instead of $k = 1 $, because the result does not change.

I have tried using "perturbation method".

Perturbation method:

$$S_{n} + x_{n+1} = x_{0} + \sum_{k=1}^{n+1} x_{k}$$

$$S_{n} + x_{n+1} = x_{0} + \sum_{k=0}^{n} x_{k+1}$$

My attempt:

$$S_{n} + (n+1) \Big( \frac{3}{2} \Big)^{n} = \sum_{k=0}^{n} (k+1)\Big(\frac{3}{2}\Big)^{k}$$

$$S_{n} + (n+1) \Big( \frac{3}{2} \Big)^{n} = \underbrace{\sum_{k=0}^{n} k \Big(\frac{3}{2}\Big)^k}_{\text{stuck here}} + \sum_{k=0}^{n} \Big(\frac{3}{2}\Big)^{k}$$

I have tried perturbing the "stuck here" sum, but it leads to infinite loop of perturbations, I think.

I'm new to sums and sequences so I don't know the tricks yet, and I think I am missing something simple to solve it.

WolframAlpha gives result:

$$\sum_{k=1}^{n} k \Big( \frac{3}{2} \Big)^{k-1} = 2 \Bigg( \Big( \frac{3}{2} \Big)^n (n-2) + 2 \Bigg)$$ Thanks for tips.

weno
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1 Answers1

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Hint: start with $$ \sum_{k=1}^n x^k = x\cdot \frac{x^n - 1}{x-1}. $$ Then differentiate it with respect to $x$: $$ \sum_{k=1}^n k x^{k-1} = \bigg( x\cdot \frac{x^n - 1}{x-1} \bigg)^{'}. $$

All you need now is to find the derivative of right hand side and substitute $x = \frac{3}{2}$

Virtuoz
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    This provides just another method but does not answer the actual question. – amsmath Sep 01 '19 at 15:56
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    The actual question was "How can I solve this sum?" I shared a hint how it can be solved. What's the problem? – Virtuoz Sep 01 '19 at 15:59
  • The problem is that you don't refer at all to the OP's attempts. – amsmath Sep 01 '19 at 16:00
  • Both your responses were helpful, I'll try to remember these tricks in the future. At the moment struggling with simplifying the derivative, but I can manage. – weno Sep 01 '19 at 16:01
  • OP said: "I'm new to sums and sequences so I don't know the tricks yet, and I think I am missing something simple to solve it". I showed this simple trick. And I still see no problem here. Anyway, thank you for downvote! – Virtuoz Sep 01 '19 at 16:06