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Question-18593




Question Number 18593 by Joel577 last updated on 25/Jul/17
Answered by 433 last updated on 25/Jul/17
      ((2016((1^(2015) /n^(2014) )+(2^(2015) /n^(2014) )...+n)−n^2 )/(2016((1^(2014) /n^(2014) )+(2^(2014) /n^(2014) )+...+(n^(2014) /n^(2014) ))))  2016n−n^2 →−∞  ((2016(0+0+...+0)−∞)/(2016(0+0+...+1)))=−∞
$$ \\ $$$$ \\ $$$$ \\ $$$$\frac{\mathrm{2016}\left(\frac{\mathrm{1}^{\mathrm{2015}} }{{n}^{\mathrm{2014}} }+\frac{\mathrm{2}^{\mathrm{2015}} }{{n}^{\mathrm{2014}} }…+{n}\right)−{n}^{\mathrm{2}} }{\mathrm{2016}\left(\frac{\mathrm{1}^{\mathrm{2014}} }{{n}^{\mathrm{2014}} }+\frac{\mathrm{2}^{\mathrm{2014}} }{{n}^{\mathrm{2014}} }+…+\frac{{n}^{\mathrm{2014}} }{{n}^{\mathrm{2014}} }\right)} \\ $$$$\mathrm{2016}{n}−{n}^{\mathrm{2}} \rightarrow−\infty \\ $$$$\frac{\mathrm{2016}\left(\mathrm{0}+\mathrm{0}+…+\mathrm{0}\right)−\infty}{\mathrm{2016}\left(\mathrm{0}+\mathrm{0}+…+\mathrm{1}\right)}=−\infty \\ $$

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