How to calculate $\sum^{n-1}_{i=0}(n-i)$?
$\sum^{n-1}_{i=0}(n-i)=n-\sum^{n-1}_{i=0}i=n-\sum^{n}_{i=1}(i-1)=2n-\frac{n(n+1)}{2}$
I am sure my steps are wrong. Could someone show me how to correct the procedure?
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1Observe that $\sum_{i=0}^{n-1} n \not= n$, so you have not pulled the $n$ out of the summation properly... – Brian Tung Apr 09 '16 at 20:33
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3Isn't this just the sum of the first n integers in backwards order? – A. Thomas Yerger Apr 09 '16 at 20:33
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@AlfredYerger: It is, but I figured OP wanted to try doing it by distributing over the summation. – Brian Tung Apr 09 '16 at 20:34
2 Answers
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Your first step is wrong:
$$\sum_{i=0}^{n-1}(n-i)=\sum_{i=0}^{n-1}n-\sum_{i=0}^{n-1}i\;,$$
and
$$\sum_{i=0}^{n-1}n=n^2\;,$$
not $n$: it’s the sum of $n$ copies of $n$.
A slicker calculation is to substitute $j=n-i$ and notice that as $i$ runs from $0$ up through $n-1$, $j$ runs from $n$ down through $1$. Since the order in which we add the terms doesn’t matter,
$$\sum_{i=0}^{n-1}(n-i)=\sum_{j=1}^nj=\frac{n(n+1)}2\;.$$
Brian M. Scott
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You have mistake in your first equation: $\sum_{0}^{n-1}{n} = n^{2}$, so you've got : $$n^{2} - \frac{n(n-1)}{2} = \frac{n(n+1)}{2}$$
openspace
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