Question

Problema 2. Fie A ={x-apartine-R/[x]*{x}=1}. Demonstrati ca:
b)daca x₁,x₂,....,x2016 ∈A si x₁<x₂<x₃<...<x2016 ,atunci   {x₁²+x₂²+...+x2016²}={x₁}²+....+{x2016}²
Indicatie suplimentara : 2016-le in toate cazurile acestui enunt este ca exponent , cum era si vizibil :)
Multumesc! :)

Answer 1

$\displaystyle Notam~k=[x]. \\ \\ \{x\}\ \textgreater \ 0 \Rightarrow k\ \textgreater ~ 0,~iar~\{x\} \neq 1 \Rightarrow k \neq 1.\\ \\ \{x\} \cdot [x]=1 \Rightarrow \{x\}= \frac{1}{k}.~Deci~x=[x]+\{x\}=k+\frac{1}{k}. \\ \\ Prin~urmare~A= \left \{ k+ \frac{1}{k} ~\Big/~k \in \mathbb{N^*}-\{1\} \right \}. \\ \\ Si~cum~x_{i}~este~element~din~A~(i= \overline{1,2016}),~putem~nota \\ \\ x_i=k_i+ \frac{1}{k_i}.$

$\displaystyle \left \{ \sum\limits^{2016}_{i=1} x_i^2 \right \}=\left \{ \sum\limits^{2016}_{i=1} \left(k_i+ \frac{1}{k_i} \right)^2 \right \}= \left\{ \sum\limits^{2016}_{i=1} \left( k_i^2+2+ \frac{1}{k_i^2}\right) \right \}= \\ \\ =\left \{ \underbrace{\sum\limits^{2016}_{i=1} \left(k_i^2+2\right)}_{\mbox{intreg}}+\sum\limits^{2016}_{i=1} \frac{1}{k_i^2} \right \}= \left \{ \sum\limits^{2016}_{i=1} \frac{1}{k_i^2} \right \}.$

$\displaystyle \sum\limits^{2016}_{i=1} \left \{ x_i \right \}^2=\sum\limits^{2016}_{i=1} \left \{\underbrace{k_i^ 2+2}_\mbox{intreg}}+ \frac{1}{k_i^2} \right \}=\sum\limits^{2016}_{i=1} \left\{ \frac{1}{k_i^2}\right\}= \sum\limits^{2016}_{i=1} \frac{1}{k_i^2}.$

$\displaystyle Din ~2 \le x_1\ \textless \ x_2\ \textless \ x_3\ \textless \ ...\ \textless \ x_{2016}~rezulta: \\ \\ \sum\limits^{2016}_{i=1} \frac{1}{k_i^2}\ \textless \ \frac{1}{2^2}+ \frac{1}{3^2}+...+ \frac{1}{2016^2}\ \textless \ \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+...+ \frac{1}{2015 \cdot2016}= \\ \\ =1- \frac{1}{2016}~(telescopare)\ \textless \ 1. \\ \\ Deci~\sum\limits^{2016}_{i=1}\frac{1}{k_i^2}= \left \{ \sum\limits^{2016}_{i=1}\frac{1}{k_i^2} \right \},~si~concluzia~rezulta.$

$Am~folosit~urmatoarele~proprietati: \\ \\ \bullet \{x+k\}=\{x\} ~\forall~ x \in \mathbb{R}~si~ k \in \mathbb{Z}. \\ \\ \bullet \{x\}=x \Leftrightarrow x \in [0;1).$