Fie \( (b_n) \) un sir de numere reale pozitive astfel incat \( b_0=1 \) si \( b_n=2+\sqrt{b_{n-1}}-2\sqrt{1+\sqrt{b_{n-1}}} \). Calculati
\( \sum_{n=1}^\infty b_n2^n \).
IMC 1995
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\( b_n = 2 + \sqrt {b_{n - 1} } - 2\sqrt {1 + \sqrt {b_{n - 1} } } = (\sqrt {1 + \sqrt {b_{n - 1} } } - 1)^2 ; \\
b_1 = (\sqrt 2 - 1)^2 ;\ b_2 = (\sqrt {1 + \sqrt {b_1 } } - 1)^2 = (\sqrt 2 - 1)^2 ; \\
\Pr {\rm in inductie completa aratam ca}\ b_k = (\sqrt 2 - 1)^2,\ \forall k \in N*\\
{\rm si}\ S = (\sqrt {\rm 2} - 1)^2 \sum\limits_{n = 1}^m {2^n } = (\sqrt 2 - 1)^2 (2^{m + 1} - 2). \)
b_1 = (\sqrt 2 - 1)^2 ;\ b_2 = (\sqrt {1 + \sqrt {b_1 } } - 1)^2 = (\sqrt 2 - 1)^2 ; \\
\Pr {\rm in inductie completa aratam ca}\ b_k = (\sqrt 2 - 1)^2,\ \forall k \in N*\\
{\rm si}\ S = (\sqrt {\rm 2} - 1)^2 \sum\limits_{n = 1}^m {2^n } = (\sqrt 2 - 1)^2 (2^{m + 1} - 2). \)
La inceput a fost numarul. El este stapanul universului.