How do you evaluate a summation with variables in the parameters?

Question

my problem is

$\sum_{i=n+1}^{3n} (2i-3)$

I have done a few summations in calc 2, but I do not remember what you are supposed to do when there are variables in both parameters.

I remember the rules that $i = \frac{(n^2+n)}{2} $ but Im not sure if that applies here

edit: adjusted the denominator which I had incorrect

If you know $$\sum_{i=1}^n i = \frac{n^2+n}{2}$$ (I fixed the denominator), see if you can write this problem in terms of that. Mainly, it tests you understanding of the $\sum$ symbol. — GEdgar, Jun 19 '20 at 19:03

Alan · Accepted Answer · 2020-06-19T19:18:58.740

0

Note that $$\sum_{i=n+1}^{3n} (2i-3) = \sum_{i=1}^{3n} (2i-3) - \sum_{i=1}^{n}(2i-3)$$ Can you take it from here?

You will have to use the fact that $$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$$

For the sake of completeness, here is the full solution:

\begin{align*}\sum_{i=n+1}^{3n} (2i-3) = \sum_{i=1}^{3n} (2i-3) - \sum_{i=1}^{n}(2i-3)&=2\sum_{i=1}^{3n}i-9n-2\sum_{i=1}^{n}i+3n \\ &=3n(3n+1)-9n-n(n+1)+3n \\ &=8n^2-4n \end{align*}

edited Jun 19 '20 at 19:18

answered Jun 19 '20 at 19:03

Alan

2,025

yes I believe I would split them up then make the change but...does the 3 from the 3n at the top just get multiplied to the whole thing? – Dingle Box Jun 19 '20 at 19:07
The $3n$ on top merely means that the upper limit of the dummy variable $i$ is $3n$. – Alan Jun 19 '20 at 19:13
do we insert that anywhere? or is that not necessary – Dingle Box Jun 19 '20 at 19:15
I have edited my answer. Does that answer your question? – Alan Jun 19 '20 at 19:19
yes perfect sense thank you so much – Dingle Box Jun 19 '20 at 19:22
You are most welcome :) – Alan Jun 19 '20 at 19:22

How do you evaluate a summation with variables in the parameters?

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