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

Answer 1

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*}