Question Number 7147 by Tawakalitu. last updated on 13/Aug/16
Commented by Yozzii last updated on 13/Aug/16
$${u}\left({n}\right)=\left(−\mathrm{1}+\mathrm{2}\right)+\left(−\mathrm{3}+\mathrm{4}\right)+\left(−\mathrm{5}+\mathrm{6}\right)+...+\left(−\left(\mathrm{2}{n}−\mathrm{1}\right)+\mathrm{2}{n}\right) \\ $$$${u}\left({n}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{2}{n}+\mathrm{1}+\mathrm{2}{n}\right)={n} \\ $$$${u}\left(\frac{\mathrm{10}^{\mathrm{9}} }{\mathrm{2}}\right)=\frac{\mathrm{10}^{\mathrm{9}} }{\mathrm{2}} \\ $$$$\mathrm{10}^{\mathrm{9}} +\underset{{k}=\mathrm{0}} {\overset{\mathrm{10}^{\mathrm{9}} } {\sum}}{k}\left(−\mathrm{1}\right)^{{k}} =\mathrm{10}^{\mathrm{9}} +\underset{{k}=\mathrm{1}} {\overset{\mathrm{10}^{\mathrm{9}} } {\sum}}{k}\left(−\mathrm{1}\right)^{{k}} =\frac{\mathrm{3}×\mathrm{10}^{\mathrm{9}} }{\mathrm{2}} \\ $$$$ \\ $$
Commented by Tawakalitu. last updated on 13/Aug/16
$${Thanks}\:{so}\:{much}\:{sir}.\:{i}\:{really}\:{appreciate} \\ $$$$ \\ $$