Sequence and series

**roshanhero** · January 13th 2010, 10:03 AM

How to do this?
$\sum_{k=1}^{100}(-1)^{k}k$

**bandedkrait** · January 13th 2010, 10:09 AM

$Sum = -1+2-3+4+........-99+100$

$=(2-1) + (4-3) +......+(100-99)$
$=1+1+1+.....+1$ (50 times)
=50??

**Drexel28** · January 13th 2010, 10:25 AM

Originally Posted by roshanhero

How to do this?
$\sum_{k=1}^{100}(-1)^{k}k$

To make a little clearer, just like bandekrait did, note that $\sum_{\ell=1}^{100}(-1)^{\ell}\ell=\sum_{\ell=1}^{50}(-1)^{2\ell+1}(2\ell-1)+\sum_{\ell=1}^{50}(-1)^{2\ell}2\ell$ (the evens in the odds) but $(-1)^{2\ell-1}=-1,(-1)^{2\ell}=1$ so this becomes $\sum_{\ell=1}^{50}-\left\{2\ell-1\right\}+\sum_{\ell=1}^{50}2\ell=\sum_{\ell=1}^{5 0}1=50$ .

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