What is the closed form of $\sum_{j=i}^{n} {j}$?

Question

How can I get a closed form from a summation like this?

$$ \sum_{j=i}^{n} {j} $$

I don’t know how to proceed since the base of the summation is a variable.

Answer 1

It's not a variable, you just start with $i$ rather than 1, so you get $i + (i+1) + \ldots n$, which is of course $\sum_{j=1}^{n}j - \sum_{j=1}^{i-1} j$. Can you handle from here?

Answer 2

You have that

\begin{align} \sum_{j=i}^n j = \sum _{j=1}^n j -\sum_{j=1}^{i-1}j=\frac{n(n+1)}{2} -\frac{i(i-1)}{2} \end{align}