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Is it possible to convert this forumla $$r(\theta) = \sum_{n=0}^\theta\left(\frac{2n+1}{2}\right)$$ to one without the $\sum$ sign? If so, how?

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up vote 3 down vote accepted

Make use of the following summations: $$\sum_{n=0}^{\theta} 1 = \theta+1$$ $$\sum_{n=0}^{\theta} n = \dfrac{\theta(\theta+1)}2$$

Hence, we have $$r(\theta) = \sum_{n=0}^{\theta}n + \dfrac12 \sum_{n=0}^{\theta} 1 = \dfrac{\theta(\theta+1)}2 + \dfrac{\theta+1}2 = \dfrac{(\theta+1)^2}2$$

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You've literally done everything for the OP. I've upvoted for your effort, but you shouldn't do everything for somebody who doesn't show an effort. Notice how my answer is detailed but still leaves some food for thought for the OP. – Ahaan S. Rungta Dec 1 '13 at 16:27

Yes. You want to use the formula for the sum of an Arithmetic Series. Note that the original sum is $$ \displaystyle\sum_{n=0}^{\theta} \dfrac {2n+1}{2} = \dfrac {1}{2} \cdot \displaystyle\sum_{n=0}^{\theta} (2n+1). $$It is easy to prove, using the formula I linked you to, that $$ \displaystyle\sum_{n=0}^{\theta} (2n+1) = (n+1)^2. $$ This is a well-known fact. Now finish up.

Edit: got beaten by 2 answers.

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