28163

From Bitnami MediaWiki
Revision as of 15:30, 16 January 2024 by Csula Beatrice (talk | contribs) (Pagină nouă: '''28163 (Dana Heuberger)''' <br /> <br /> ''Aflați șirul de numere naturale nenule <math>(a_n)_{n\geq1}</math> pentru care <math>\frac{1}{{(1+a_1) \cdot a_{a_1}}} + \frac{1}{{(1+a_2) \cdot a_{a_2}}} + \ldots + \frac{1}{{(1+a_n) \cdot a_{a_n}}} = \frac{n}{{n+1}}</math>, pentru orice <math>n \geq 1</math>.'' <br /> <br /> '''Soluție:''' Dacă <math>n = 1</math>, egalitatea din enunț devine <math>(1+a_1) \cdot a_{a_1} = 2</math>, de unde obținem <math>a_1 = 1</math>. </b...)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

28163 (Dana Heuberger)

Aflați șirul de numere naturale nenule 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)_{n\geq1}} pentru care 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{1}{{(1+a_1) \cdot a_{a_1}}} + \frac{1}{{(1+a_2) \cdot a_{a_2}}} + \ldots + \frac{1}{{(1+a_n) \cdot a_{a_n}}} = \frac{n}{{n+1}}} , 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 \geq 1} .

Soluție:

Dacă 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 = 1} , egalitatea din enunț devine 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 (1+a_1) \cdot a_{a_1} = 2} , de unde 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_1 = 1} .

Dacă 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 \geq 2} , scriind egalitatea din enunț pentru 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-1} și scăzând-o din relația inițială, 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{1}{{(1+a_n) \cdot a_{a_n}}} = \frac{n}{{n+1}} - \frac{n-1}{n} = \frac{1}{{n(n+1)}}} , deci

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 (1+a_n) \cdot a_{a_n} = n(n+1)} 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 \in \mathbb{N}} , 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 \geq 2} . (1)

Demonstrăm, folosind inducția tare, că 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 = n} , pentru orice număr natural nenul 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} . Etapa de verificare este evidentă.
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 k \in \mathbb{N}} , 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 \geq 2} . Presupunem că 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_t = t} , pentru orice număr natural 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 t} cu 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 t \in \{1,2,...,k-1\}} și arătăm că 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_k = k} .
I. Dacă 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_k = s < k} , din ipoteza de inducție 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 a_{a_k} = a_s = s < k} . Din relația (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 k(k+1) = (1+a_k) \cdot a_{a_k} = s(s+1) < k(k+1)} , fals.

II. Dacă 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_k = s > k} , din (1) deducem:
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_s = a_{a_k} = \frac{k(k+1)}{1+a_k} = \frac{k(k+1)}{s+1}, } deci 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_s = k \cdot \frac{k+1}{s+1} < k} . (2)

Din ipoteza de inducție rezultă că 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_{a_s} = a_s = \frac{k(k+1)}{s+1}} .
Din relația (1) obținem că 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_{a_s} = \frac{s(s+1)}{a_s+1}} , deci 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{k(k+1)}{s+1} = \frac{s(s+1)}{1+a_s}} .
Așadar, 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 1+a_s = \frac{s(s+1)}{k(k+1)} \cdot (s+1) > s+1} , adică 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_s > s > k} , contradicție cu inegalitatea (2).
Din I și II deducem că 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_k = k} . Conform principiului inducției, rezultă că 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 = n} , pentru orice număr natural 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} . Pentru acest șir, egalitatea din enunț devine o identitate, așadar soluția problemei este șirul cu termenul general 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 = n} .




Retrieved from "http://wiki.universitas.ro/index.php?title=28163&oldid=9523"