28250

_{28250 (Codruț-Sorin Zmicală)}

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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} .