Question

Daca $\frac{6}{31}$ = 0,$a_{1}$$a_{2}$$a_{3}$...$a_{n}$...

Sa se calculeze: $a_{53}$ + $a_{54}$ + ... + $a_{80}$

Answer 1

$\frac{6}{31}= 0,(193548387096774) ~~~~~~~(dificil~calcul...)\\ \\ Deci~numarul~ \frac{6}{31}~are~perioada~de~lungime~15. ~Asadar~a_n=a_{n+15k}. \\ \\(n,k \in N^*). \\ \\ 53=8+15 \cdot 3 \\ \\54=9+15 \cdot 3 \\ \\ 55=10+15 \cdot 3 \\ \\.........................$

$59=14 + 3 \cdot 15 \\ \\ 60=4 \cdot 15 \\ \\ ......................... \\ \\ Deci~a_{53}=a_{8}=8,~a_{54}=a_9=7,~a_{55}=a_{10}=0,... \\ \\ S=(8+7+0+9+6+7+7+4)+(1+9+3+...+7+4)+\\ \\ +(1+9+3+5+4)=151.$