# Sequence and series

• January 13th 2010, 11:03 AM
roshanhero
Sequence and series
How to do this?
$\sum_{k=1}^{100}(-1)^{k}k$
• January 13th 2010, 11:09 AM
bandedkrait
$Sum = -1+2-3+4+........-99+100$

$=(2-1) + (4-3) +......+(100-99)$
$=1+1+1+.....+1$(50 times)
=50??
• January 13th 2010, 11:25 AM
Drexel28
Quote:

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 $\sum_{\ell=1}^{50}-\left\{2\ell-1\right\}+\sum_{\ell=1}^{50}2\ell=\sum_{\ell=1}^{5 0}1=50$.