Subiectul II Gheorghe Lazar 2009 Sibiu

[DrAGos Calinescu]

Fie \( {(a_k)}_{k\ge 1} \) o progresie aritmetica cu ratia strict pozitiva \( r \) si primul termen \( a_1\ge\frac{1}{2} \).
Calculati partea intreaga a numarului
\( s_n=\sqrt{1+\frac{r}{a_1a_2}}+\sqrt{1+\frac{r}{a_2a_3}}+\dots +\sqrt{1+\frac{r}{a_{n-1}a_n}}+\sqrt{1+\frac{1}{a_n}} \), pentru orice \( n\in\mathbb{N} ,n\ge 2. \)

Dumitru Acu, Sibiu

[Beniamin Bogosel]

Eu zic ca un CBS clarifica situatia ...

[Virgil Nicula]

Fie \( {(a_k)}_{k\ge 1} \) o progresie aritmetica cu ratia \( r>0 \) si \( a_1\ge\frac{1}{2} \) . Calculati partea intreaga a numarului

\( s_n=\sqrt{1+\frac{r}{a_1a_2}}+\sqrt{1+\frac{r}{a_2a_3}}+\dots +\sqrt{1+\frac{r}{a_{n-1}a_n}}+\sqrt{1+\frac{1}{a_n}} \), pentru orice \( n\in\mathbb{N}\ ,\ n\ge 2. \) .

Dem. \( n\ <\ s_n\stackrel{(C.B.S.)}{\ \ \ \ <\ \ \ \ }\ \sqrt {n\cdot\left[n+\frac {1}{a_n}+\sum_{k=1}^{n-1}\left(\frac {1}{a_k}-\frac {1}{a_{k+1}}\right)\right]}\ =\ \sqrt {n\left(n+\frac {1}{a_1}\right)}\ \le\ \sqrt {n(n+2)}\ <\ n+1\ . \)
In concluzie \( (\forall )\ n\ge 2\ ,\ \left[s_n\right]=n \) . Am folosit inegalitatea Cebasev \( \left(\sum_{k=1}^nx_k\right)^2\ \le\ n\cdot\sum_{k=1}^nx^2_k \) .