Demonstrati ca \( \sqrt{n^2+1}+\sqrt {n^2+2}+.....+\sqrt{n^2+2n}<\frac{4n^2+2n+1}{2} \),pentru orice \( n\in N* \).
Lucian Tutescu, Craiova, Recreatii Matematice 1/2008
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Andi Brojbeanu
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Marius Mainea
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Se foloseste AM-GM.
\( \sqrt{n^2+k}=n\sqrt{(1+\frac{k}{n^2})\cdot1}\le n\cdot\frac{1+\frac{k}{n^2}+1}{2}=n+\frac{k}{2n} \)
\( \sqrt{n^2+k}=n\sqrt{(1+\frac{k}{n^2})\cdot1}\le n\cdot\frac{1+\frac{k}{n^2}+1}{2}=n+\frac{k}{2n} \)
Last edited by Marius Mainea on Mon Jun 29, 2009 10:15 pm, edited 1 time in total.
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Andi Brojbeanu
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Demonstram ca \( \sqrt{n^2+k}<n+\frac{k}{2n} \). Ridicand la patrat, obtinem \( n^2+k<n^2+k+ \frac{k^2}{4n^2} \), adevarat.
Inlocuind in relatia diin cerinta, avem \( \sqrt{n^2+1}+\sqrt{n^2+2}+....+\sqrt{n^2+2n}<n+\frac{1}{2n}+n+\frac{2}{2n}+....+n+\frac{2n}{2n}=2n^2+\frac{\frac{2n\cdot(2n+1)}{2}}{2n}=\frac{4n^2+2n+1}{2} \), adica ceea ce trebuia demonstrat.
Inlocuind in relatia diin cerinta, avem \( \sqrt{n^2+1}+\sqrt{n^2+2}+....+\sqrt{n^2+2n}<n+\frac{1}{2n}+n+\frac{2}{2n}+....+n+\frac{2n}{2n}=2n^2+\frac{\frac{2n\cdot(2n+1)}{2}}{2n}=\frac{4n^2+2n+1}{2} \), adica ceea ce trebuia demonstrat.