28163

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 $n=1$, egalitatea din enunț devine $(1+a_{1})\cdot a_{a_{1}}=2$, de unde obținem $a_{1}=1$.

Dacă $n\geq 2$, scriind egalitatea din enunț pentru $n-1$ și scăzând-o din relația inițială, obținem

${\frac {1}{(1+a_{n})\cdot a_{a_{n}}}}={\frac {n}{n+1}}-{\frac {n-1}{n}}={\frac {1}{n(n+1)}}$, deci

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

Demonstrăm, folosind inducția tare, că $a_{n}=n$, pentru orice număr natural nenul $n$. Etapa de verificare este evidentă.
Fie $k\in \mathbb {N}$, $k\geq 2$. Presupunem că $a_{t}=t$, pentru orice număr natural $t$ cu $t\in \{1,2,...,k-1\}$ și arătăm că $a_{k}=k$.
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} .