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Amid one exercise I was solving, I came across the following finite sum:

$$ \sum_{n=0}^{N} n\left(\frac{3}{2}\right)^n.$$

This sum was evaluated in one of my classes, but I don't understand/agree with the resolution provided. For context, I present the resolution presented in class:

$$ \sum_{n=0}^N n\left(\frac{3}{2}\right)^n = \sum_{n=0}^N n\left(\frac{3}{2}\right)^n = \frac{3}{2} \sum_{n=0}^N n\left(\frac{3}{2}\right)^{n-1} \color{red}{= \frac{3}{2} \sum_{n=0}^N \left( \left( \frac{3}{2} \right)^n \right)'} = \frac{3}{2} \left( \sum_{n=0}^N \left(\frac{3}{2}\right)^n \right) ' = \frac{3}{2} \left( \frac{1.5^{N+1} - 1}{1.5-1} \right)'.$$

In the step I colored in red, it seems like we took the derivative of $(1.5)^n$ with respect to $1.5$, which I don't know if is possible. In the case that such computation makes sense, then everything that follows is clear, but I don't think derivatives with respect to a constant make sense.

Thus, I am looking for an alternative way of evaluating this sum.

Thanks for any help in advance.

xyz
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  • https://math.stackexchange.com/q/3145414/42969, https://math.stackexchange.com/q/119636/42969, https://math.stackexchange.com/q/180198/42969 – Martin R Dec 09 '23 at 12:24
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    That's a terrible way to write a sensible operation. I think the intent becomes clear if you replace $\frac 32$ with $x$ and differentiate with respect to $x$. You can then evaluate at $x=\frac 32$. – lulu Dec 09 '23 at 12:26
  • You can't derivate with respect to a number. One derivates with respect to variables. – jjagmath Dec 09 '23 at 12:50

1 Answers1

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Consider $$\sum_{n=0}^N x^n $$ derive with respect to $x$ and then plug $x=3/2$.

Sine of the Time
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