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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.

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    $\begingroup$ Agreed. And it should calculate nicely. $\endgroup$ – gnometorule Feb 19 '13 at 3:32
  • $\begingroup$ @gnometorule Thanks! It's been a while since I summed anything. $\endgroup$ – TheHopefulActuary Feb 19 '13 at 3:47
  • $\begingroup$ So you know how to compute this sum, right? Or are you asking how to do it? $\endgroup$ – Julien Feb 19 '13 at 3:51
  • $\begingroup$ Yes I knew how to do it as long as what I had was the right setup. I small part of me doubted myself for some reason. $\endgroup$ – TheHopefulActuary Feb 19 '13 at 3:56
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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 $\sum (2n+1)=2\sum n +\sum 1$.

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