TITLE: Did I sum this correctly?
QUESTION [5 upvotes]: I was tutoring some students in calculus today and they were supposed to sum up all of the odd four digit numbers. They weren't really confident about their answers so I showed them what I would do. Which is this: $$\sum_{n=500}^{4999}2n+1$$ If I understood the question correctly they were supposed to sum up $1001+1003+1005+...+9999$, is that correct? I feel confident but I have a little doubt in me.

REPLY [6 votes]: All you really need to know to compute this is the formula
$$
\sum_{n=1}^Nn=\frac{N(N+1)}{2}.
$$
Also, note that here you have $\sum_a^b=\sum_1^b-\sum_1^{a-1}$.
And that $\sum (2n+1)=2\sum n +\sum 1$.