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As stated above, the series is:

1 + 2 - 3 + 4 - 5 + 6 - 7 + 8 - 9 + 10 - 11 + 12 - 13 + 14 - 15 . . . ± (n - 1) ∓ n

What would the sigma notation be for this series, starting at 1, ending at n?

Thank you very much for your time!

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$$2+\sum_{k=1}^n (-1)^{k}\cdot k$$

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I don't understand the downvote, either. If the first term is +1, one can just add 2 to this. – Ross Millikan Oct 3 '12 at 3:21
    
Nah. Meant it to start with the +1. – Jessica Hope Oct 3 '12 at 3:22
    
@JessicaHope Adjusted. – user39572 Oct 3 '12 at 3:23
    
Thank you very much, Julien! – Jessica Hope Oct 3 '12 at 3:28

i think $$\sum_{k=1}^{n}(-1)^{k+\lfloor\frac{1}{k}\rfloor}\cdot k$$ works

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A more abstruse solution: $$\sum_{k=1}^n (-1)^{k \cdot \textrm{sgn} (k-1)}k$$ where $\textrm {sgn}$ is the signum function.

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