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28250 - Revision history
2026-06-16T22:53:22Z
Revision history for this page on the wiki
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https://wiki.universitas.ro/index.php?title=28250&diff=7108&oldid=prev
Andrei.Horvat at 12:06, 31 October 2023
2023-10-31T12:06:06Z
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<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 12:06, 31 October 2023</td>
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<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>''Calculați''</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>''Calculați''</div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>''<math>\lim_{n \to \infty}\sqrt[n]{\int_{0}^{1} (\sqrt{x}+x^n})^ndx</math><del style="font-weight: bold; text-decoration: none;">.</del>''</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>''<math <ins style="font-weight: bold; text-decoration: none;">display="block"</ins>>\lim_{n \to \infty}\sqrt[n]{\int_{0}^{1} (\sqrt{x}+x^n})^ndx<ins style="font-weight: bold; text-decoration: none;">.</ins></math><ins style="font-weight: bold; text-decoration: none;">'''''Soluție:'</ins>''</div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">'''Soluție:'''</del></div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Fie <math>a_n=\int_{0}^{1} (\sqrt{x}+x^n)^ndx</math>, n<math>\in\Nu^*</math>. Cu binomul lui Newton avem <math>(\sqrt{x}+x^n)^n=\sum_{k=0}^n\binom{n}{k}x^\tfrac{(2n-1)k+n}{2}</math>, iar prin integrare pe [0,1] obținem <math <ins style="font-weight: bold; text-decoration: none;">display="block"</ins>>a_n=\sum_{k=0}^n\binom{n}{k}\cdot\frac{2}{(2n-1)k+n+2}<ins style="font-weight: bold; text-decoration: none;">.</ins></math>Pentru orice <math>k\in\{0,1,...,n\}</math> avem <math <ins style="font-weight: bold; text-decoration: none;">display="block"</ins>>\binom{n}{k}\cdot\frac{1}{n^2+1}\leqslant\binom{n}{k}\cdot\frac{2}{(2n-1)k+n+2}\leqslant\binom{n}{k}<ins style="font-weight: bold; text-decoration: none;">,</ins></math>iar prin însumarea acestor inegalități obținem <math <ins style="font-weight: bold; text-decoration: none;">display="block"</ins>>\frac{2^n}{n^2+1}\leqslant a_n\leqslant 2^n<ins style="font-weight: bold; text-decoration: none;">.</ins></math>Rezultă <math>\frac{2}{\sqrt[n]{n^2+1}}\leqslant{\sqrt[n]{a_n}}\leqslant2</math>, pentru orice <math>n\geqslant2</math>. Cum <math>\lim_{n \to \infty}\sqrt[n]{n^2+1}=1</math>, din teorema cleștelui obținem <math <ins style="font-weight: bold; text-decoration: none;">display="block"</ins>>\lim_{n \to \infty}\sqrt[n]{a_n}=2<ins style="font-weight: bold; text-decoration: none;">.</ins></math></div></td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div> </div></td><td colspan="2" class="diff-side-added"></td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Fie <math>a_n=\int_{0}^{1} (\sqrt{x}+x^n)^ndx</math>, n<math>\in\Nu^*</math>. Cu binomul lui Newton avem <math>(\sqrt{x}+x^n)^n=\sum_{k=0}^n\binom{n}{k}x^\tfrac{(2n-1)k+n}{2}</math>, iar prin integrare pe [0,1] obținem <math>a_n=\sum_{k=0}^n\binom{n}{k}\cdot\frac{2}{(2n-1)k+n+2}</math><del style="font-weight: bold; text-decoration: none;">. </del></div></td><td colspan="2" class="diff-side-added"></td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div> </div></td><td colspan="2" class="diff-side-added"></td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Pentru orice <math>k\in\{0,1,...,n\}</math> avem <math>\binom{n}{k}\cdot\frac{1}{n^2+1}\leqslant\binom{n}{k}\cdot\frac{2}{(2n-1)k+n+2}\leqslant\binom{n}{k}</math><del style="font-weight: bold; text-decoration: none;">, </del>iar prin însumarea acestor inegalități obținem <math>\frac{2^n}{n^2+1}\leqslant a_n\leqslant 2^n</math><del style="font-weight: bold; text-decoration: none;">. </del></div></td><td colspan="2" class="diff-side-added"></td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div> </div></td><td colspan="2" class="diff-side-added"></td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Rezultă <math>\frac{2}{\sqrt[n]{n^2+1}}\leqslant{\sqrt[n]{a_n}}\leqslant2</math>, pentru orice <math>n\geqslant2</math>. Cum <math>\lim_{n \to \infty}\sqrt[n]{n^2+1}=1</math>, din teorema cleștelui obținem <math>\lim_{n \to \infty}\sqrt[n]{a_n}=2</math><del style="font-weight: bold; text-decoration: none;">.</del></div></td><td colspan="2" class="diff-side-added"></td></tr>
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Andrei.Horvat
https://wiki.universitas.ro/index.php?title=28250&diff=7070&oldid=prev
Ghetie Gabriela Claudia: Pagină nouă: <sub>'''<big>28250 (Codruț-Sorin Zmicală)</big>'''</sub> ''Calculați'' ''<math>\lim_{n \to \infty}\sqrt[n]{\int_{0}^{1} (\sqrt{x}+x^n})^ndx</math>.'' '''Soluție:''' Fie <math>a_n=\int_{0}^{1} (\sqrt{x}+x^n)^ndx</math>, n<math>\in\Nu^*</math>. Cu binomul lui Newton avem <math>(\sqrt{x}+x^n)^n=\sum_{k=0}^n\binom{n}{k}x^\tfrac{(2n-1)k+n}{2}</math>, iar prin integrare pe [0,1] obținem <math>a_n=\sum_{k=0}^n\binom{n}{k}\cdot\frac{2}{(2n-1)k+n+2}</math>. Pentru orice <ma...
2023-10-28T14:07:48Z
<p>Pagină nouă: <sub>'''<big>28250 (Codruț-Sorin Zmicală)</big>'''</sub> ''Calculați'' ''<math>\lim_{n \to \infty}\sqrt[n]{\int_{0}^{1} (\sqrt{x}+x^n})^ndx</math>.'' '''Soluție:''' Fie <math>a_n=\int_{0}^{1} (\sqrt{x}+x^n)^ndx</math>, n<math>\in\Nu^*</math>. Cu binomul lui Newton avem <math>(\sqrt{x}+x^n)^n=\sum_{k=0}^n\binom{n}{k}x^\tfrac{(2n-1)k+n}{2}</math>, iar prin integrare pe [0,1] obținem <math>a_n=\sum_{k=0}^n\binom{n}{k}\cdot\frac{2}{(2n-1)k+n+2}</math>. Pentru orice <ma...</p>
<p><b>New page</b></p><div><sub>'''<big>28250 (Codruț-Sorin Zmicală)</big>'''</sub><br />
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''Calculați''<br />
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''<math>\lim_{n \to \infty}\sqrt[n]{\int_{0}^{1} (\sqrt{x}+x^n})^ndx</math>.''<br />
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'''Soluție:'''<br />
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Fie <math>a_n=\int_{0}^{1} (\sqrt{x}+x^n)^ndx</math>, n<math>\in\Nu^*</math>. Cu binomul lui Newton avem <math>(\sqrt{x}+x^n)^n=\sum_{k=0}^n\binom{n}{k}x^\tfrac{(2n-1)k+n}{2}</math>, iar prin integrare pe [0,1] obținem <math>a_n=\sum_{k=0}^n\binom{n}{k}\cdot\frac{2}{(2n-1)k+n+2}</math>. <br />
<br />
Pentru orice <math>k\in\{0,1,...,n\}</math> avem <math>\binom{n}{k}\cdot\frac{1}{n^2+1}\leqslant\binom{n}{k}\cdot\frac{2}{(2n-1)k+n+2}\leqslant\binom{n}{k}</math>, iar prin însumarea acestor inegalități obținem <math>\frac{2^n}{n^2+1}\leqslant a_n\leqslant 2^n</math>. <br />
<br />
Rezultă <math>\frac{2}{\sqrt[n]{n^2+1}}\leqslant{\sqrt[n]{a_n}}\leqslant2</math>, pentru orice <math>n\geqslant2</math>. Cum <math>\lim_{n \to \infty}\sqrt[n]{n^2+1}=1</math>, din teorema cleștelui obținem <math>\lim_{n \to \infty}\sqrt[n]{a_n}=2</math>.</div>
Ghetie Gabriela Claudia