How do I evaluate a series?

Question

In this specific example, I don't understand the steps of evaluating this series: \begin{align} &\frac{12}{n}\left(\left[\sum_{i=1}^n-7\right]+\sum_{i=1}^n\left[\frac{-12}{n}\cdot i\right]\right)\\ &=\frac{-84}{n}\sum_{i=1}^n(1)+\frac{-144}{n^2}\cdot\sum_{i=1}^n(i)\\ &=\frac{-84}{n}\cdot n+\frac{-144}{n^2}\cdot\frac{n\cdot (n+1)}{2} \end{align}

Answer 1

It ultimately relies on the two observations $$\underbrace{c+c+\cdots+c}_{n\textrm{ times}} =n\cdot c$$ and $$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$$

You should memorize these formulas, they are used rather frequently. Note that the summation notation disguises them somewhat. They could be written as

$$\sum_{k=1}^n c =n\cdot c$$ $$\sum_{k=1}^n k =\frac{n(n+1)}{2}$$

(and of course the name "$k$" of the internal dummy index is irrelevant).

Answer 2

For the first series:

$$\frac{12}{n}\sum_{i=1}^n -7=\frac{12}{n}\sum_{i=1}^n -7\cdot1=\frac{12}{n}\cdot(-7)\sum_{i=1}^n 1=\frac{-84}{n}\sum_{i=1}^n 1$$

Similar for the second one. Then, use the fact that

$$\sum_{i=a}^b 1=(b-a+1)$$

and

$$\sum_{i=a}^b i=\frac{(a+b)(b-a+1)}{2}$$

for $a=1$ and $b=n$.