Double Summation Simplification (Simple) 
My questions:


*

*For the inner summation, When $j=i$ does that mean $j$ will equal $2$ because $i$ equals $2$? 

*I'm confused and stuck on the yellow, because I thought it would just be $n$.  NOT ($(n-i+n)$. I understandstand Line 1 of the problem somewhat. It's just the index of the inner j=i is throwing me off. 
 A: The meaning of summation notation $ \Sigma $ follows as:
$$ \sum^{n}_{k=i}(\text{formula of }k) = \text{Let's sum a formula of }k\text{  when }k=i, i+1, i+2 \ldots n.$$
so for your question 1, j=i does not mean j=2, even if it is placed right after i=2.
Actually,
$$ \sum^{n}_{k=i}(\text{formula of }k)$$
can be written equivalently as(but don't use this notation if your lecture giver insists)
$$ \sum_{k \in \{i, i+1, i+2, ... n\}}(\text{formula of }k)$$
so an equal sign does not have the meaning you think.
So, for your question 2, 
$$ \sum^{n}_{j=i}1 = \sum_{j \in \{i, i+1, i+2 ... n\}} 1 \\
= \text{sum of }(n - i + 1) \text{ } 1\text{'s} \\
= n - i + 1.$$
Hope my answer helps.
A: For each $2 \leq i \leq n$, we have 
$$
\sum_{j=i}^{n}3i = \underset{(n-i+1\ \text{times})}{3i + \cdots + 3i} = (n-i+1)3i.
$$
For each $i \geq 1$
the number of the elements of the set $\{ i, i+1, \dots, n \}$ is simply $n-i+1$, as you can try to prove; for instance, if $i=2$ and if $n = 4$, then the set $\{ i, i+1, \dots, n\} = \{ 2, 3, 4 \}$ has $3 = n-i+1$ elements.
