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$E_{n}=1000$( (n + 1) {*} $\frac{n}{2}$ )

$E{}_{n}=500((n+1)*n)$

$E_{n}=500(n^{2}+n)=500n^{2}+500n$

$0=500n^{2}+500n-E_{n}$

nach der bekannten Formel zur Bestimmung quadratischer Gleichungen:

$n_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-500n\pm\sqrt{(500n)^{2}-4*500n^{2}*(-E_{n})}}{2*500n^{2}}=\frac{-500n\pm\sqrt{250000n{}^{2}+2000n^{2}E_{n}}}{2*500n^{2}}=$

$=\frac{-500n\pm20n\sqrt{625+5E_{n}}}{1000n^{2}}=\frac{-25\pm\sqrt{625+5E_{n}}}{50n}$
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