28250: Difference between revisions

Latest revision as of 12:06, 31 October 2023

_{28250 (Codruț-Sorin Zmicală)}

Calculați

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty}\sqrt[n]{\int_{0}^{1} (\sqrt{x}+x^n})^ndx.} Soluție:

Fie Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_n=\int_{0}^{1} (\sqrt{x}+x^n)^ndx} , nFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \in\Nu^*} . Cu binomul lui Newton avem Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\sqrt{x}+x^n)^n=\sum_{k=0}^n\binom{n}{k}x^\tfrac{(2n-1)k+n}{2}} , iar prin integrare pe [0,1] obținem Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_n=\sum_{k=0}^n\binom{n}{k}\cdot\frac{2}{(2n-1)k+n+2}.} Pentru orice Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k\in\{0,1,...,n\}} avem Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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},} iar prin însumarea acestor inegalități obținem Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2^n}{n^2+1}\leqslant a_n\leqslant 2^n.} Rezultă Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2}{\sqrt[n]{n^2+1}}\leqslant{\sqrt[n]{a_n}}\leqslant2} , pentru orice Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n\geqslant2} . Cum Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty}\sqrt[n]{n^2+1}=1} , din teorema cleștelui obținem Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty}\sqrt[n]{a_n}=2.}

@@ Line 3: / Line 3: @@
 ''Calculați''
-''<math>\lim_{n \to \infty}\sqrt[n]{\int_{0}^{1} (\sqrt{x}+x^n})^ndx</math>.''
+''<math display="block">\lim_{n \to \infty}\sqrt[n]{\int_{0}^{1} (\sqrt{x}+x^n})^ndx.</math>'''''Soluție:'''
-'''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 display="block">a_n=\sum_{k=0}^n\binom{n}{k}\cdot\frac{2}{(2n-1)k+n+2}.</math>Pentru orice <math>k\in\{0,1,...,n\}</math> avem <math display="block">\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 display="block">\frac{2^n}{n^2+1}\leqslant a_n\leqslant 2^n.</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 display="block">\lim_{n \to \infty}\sqrt[n]{a_n}=2.</math>
-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 <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>.
-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>.