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The context is as follows: I am asking this question because I would like feedback; I am a beginner to mathematical proofs.

We wish to show $\sum\limits_{k=1}^{n}kq^{k-1} = \frac{1-(n+1)q^{n} + nq^{n+1}}{(1-q)^{2}}$ for arbitrary $q>0$. My attempt:

Let as first consider the trivial base case $n=1$ which, in the series, is clearly equally to one. Upon simplifying the left side using basic algebra, we see that the fraction is also equal to one. The base case has been shown.

Now, we must show the inductive step, meaning that if the equality holds for some $n$ it necessarily holds for $n+1$. Let $r=n+1$. We add to the series the next term $r*q^{r-1} = (n+1)q^{n}$ (to both sides of the equation). Thus,

$\frac{1-(n+1)q^{n} + nq^{n+1}}{(1-q)^{2}} + (n+1)q^{n}$ $\Rightarrow$ $\frac{1-(n+1)q^{n} + nq^{n+1} + (1-q)^2(n+1)(q^n)}{(1-q)^{2}}$

$\Rightarrow$ $\frac{1-(n+1)q^{n} + nq^{n+1} + (n+1)(q^n - 2q^{n+1} + q^{n+2})}{(1-q)^{2}}$ $\Rightarrow$ $\frac{1 + nq^{n+1} + (n+1)(q^{n+2} - 2q^{n+1})}{(1-q)^{2}}$ $= \frac{1 - (n+2)q^{n+1} + (n+1)(q^{n+2})}{(1-q)^{2}}$

Substituting $r=n+1$ the final expression can be simplified to $\frac{1 - (r+1)q^{r} + (r)(q^{r+1})}{(1-q)^{2}}$

Therefore it's clear that if the original equality is true for some number $n$, it is also true for the number $r=n+1$ and this process can be repeated ad infinitum. The "first" case was the base case and the rest follows logically.

Insert Pseudonym
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2 Answers2

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The finite geometric series is given by

$$ G(q,n)=\sum_{k=0}^nq^k=\frac{1-q^{n+1}}{1-q} $$ for constant $q$ such that $|q|<1$.

Now $\frac{d}{dq}G(q,n)$ is the series you are looking for...

JejeBelfort
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tired
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Start with the standard formula for the sum of the first $n$ terms of a geometric series and differentiate both sides. This will give you the formula you have

David Quinn
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