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There is an exercise with the following notation:

$$ \sum_{i=0}^{2n}3x^{-n} $$

Is this a correct notation?

According to the book the result of the calculation is $ 3(2n+1)x^{-n} $

How is it possible?

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    Usually, one would use $i$ in the exponent because $n$ is fixed in this problem. – 2017-02-02
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    Take the $3$ on the outside. Now use the formula $$1 + a + \cdots + a^r = \frac{1 - a^{r+1}}{1-a}$$ – 2017-02-02
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    I don't know how a geometric series can be correct. – 2017-02-02
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    It can be calculated in the way you wrote it but it's unusual to write it like this and is most likely a typo, where it was supposed to say $\sum_{i=0}^{2n}3x^{-i}$. – 2017-02-02
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    @D_S: The OP is asking whether the expression is well-formed. As Michael Burr mentions, the summand should probably be $3x^{-i}$ or something like that—at any rate, probably not dependent on $n$, since that is not the index of summation. Then it can be summed, using the formula you cite. – 2017-02-02
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    @BrianTung It should be made clear, though, that the expression *is* well-formed as it is too. It's just uncommon to write it like this. – 2017-02-02
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    This is almost certainly a typo. In the most literal sense, however, it *can* be computed: since the value of the expression doesn't vary with the value of $i$, it would simply reduce to: $$6nx^{-n}$$ – 2017-02-02
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    @Wildcard You disappointed me! There are $2n+1$ numbers from $0$ to $2n$, inclusively. :D – 2017-02-02
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    @TimonG.: You are quite right. I sort of skipped right past that in my comment. – 2017-02-02
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    @SimplyBeautifulArt, D'oh! $(6n+3)x^{-n}$, then – 2017-02-02

2 Answers 2

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The calculation is correct, since we obtain \begin{align*} \sum_{i=0}^{2n}3x^{-n}=3x^{-n}\sum_{i=0}^{2n}1=3x^{-n}(2n+1) \end{align*}

Here we use that $3x^{-n}$ does not depend on the summation index $i$ and we can apply the distributive law \begin{align*} \sum_{i=0}^{2n}3x^{-n}&=\sum_{i=0}^{2n}\left(3x^{-n}\cdot 1\right)=3x^{-n}\sum_{i=0}^{2n}1 \end{align*} as well as \begin{align*} \sum_{i=0}^{a}1=a+1 \end{align*}

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    Yes of course! Thanks! – 2017-02-03
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    @Qualid: You're welcome! :-) – 2017-02-03
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The notation is valid, but it's probably not what you expect. To break down what's happening:

We are taking the sum from $i = 0$ to $i = 2n$ of some sequence $a_i$. In other words, we will look at the terms of the sequence $a_0, a_1, \ldots, a_{2n-1}, a_{2n}$ and add them together. There are a grand total of $2n+1$ terms.

The sequence is defined as $a_i = 3x^{-n}$. But $i$ doesn't actually appear in the expression, which means that all terms of the sequence are equal. The sequence is $\{3x^{-n}, 3x^{-n}, 3x^{-n}, \ldots\}$. So we are adding together $2n+1$ copies of $3x^{-n}$, so the total of the series is $(2n+1) \times 3x^{-n}$.

Now, it seems likely that there was a typo, and the intended sum was $\sum_{i=0}^{2n} 3x^{-i}$. In this case, the series looks like this (starting with the 0th term): $\{3, 3x^{-1}, 3x^{-2}, 3x^{-3}, \ldots\}$. So you start at $3$ and each successive term divides the previous one by $x$, or equivalently multiplies it by $x^{-1}$.

Then this is a geometric series with $2n+1$ terms, first term $3$ and ratio $x^{-1}$, and the sum is $3\left(\frac{1-x^{-(2n+1)}}{1-x^{-1}}\right)=3\frac{x-x^{-2n}}{x-1}$.

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    Interestingly, even with the typo (if that's what it is), formula denotes a geometric series; it's just a very boring series with ratio $1.$ – 2017-02-02
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    True! Although one could also argue it's an arithmetic series with difference 0. – 2017-02-03
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    In fact it's both! :-) – 2017-02-03
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    @ConMan you are correct in your reasoning. – 2017-02-03