Cristian Calude, proba pe echipe, R.II, P.III

[Laurian Filip]

Sa se determine numerele \( a_1,a_2,a_3,...,a_n \) astfel incat
\( \sqrt{a_1 }+\sqrt{a_2-1}+\sqrt{a_3-2}+...+\sqrt{a_n-(n-1)}=\frac{1}{2}(a_1+a_2+...+a_n)-\frac{n(n-3)}{4} \)

[alex2008]

Facem substitutia :

\(
\sqrt {a_1} = b_1 \\
\sqrt {a_2 - 1} = b_2 \\
\sqrt {a_3 - 2} = b_3 \\
\vdots \\
\sqrt {a_n - (n - 1)} = b_n \)

Deci avem :

\( \sum_{i = 1}^{n}b_i = \frac {1}{2}\left(b_1^2 + b_2^2 + 1 + b_3^2 + 2 + \ldots + b_n^2 + n - 1 \right) - \frac {n(n - 3)}{4} \)

\( 2\cdot \sum_{i = 1}^{n}b_i - \sum_{i = 1}^{n}b_i^2 = \frac {n(n - 1) - n(n - 3)}{2} = n \)

\( \sum_{i = 1}^{n}b_i^2 - 2\cdot \sum_{i = 1}^{n}b_i + \sum_{i = 1}^{n}1 = 0 \)

\( \sum_{i = 1}^{n}(b_i - 1)^2 = 0 \)

Deci :
\(
b_1 - 1 = \sqrt {a_1} - 1 = 0 \Rightarrow a_1 = 1 \\
b_2 - 1 = \sqrt {a_2 - 1} - 1 = 0\Rightarrow a_2 = 2 \\
b_3 - 1 = \sqrt {a_3 - 2} - 1 = 0\Rightarrow a_3 = 3 \\
\vdots \)

Deci \( a_n = n \)