Question

Fie numarul:
a=$\frac{1}{1 ori 2}+ \frac{1}{2 ori 3} +...+ \frac{1}{n(n+1)} ,$,n≥1
Sa se determine n apartine N* astfel incat {a}=0,999

Answer 1

1/1*2=1-1/2
1/2*3=1/2-1/3
...............................
1/.n*(n+1)=1/n-1/(n+1)
________________ se  aduna  termen  cutermen 
1/1*2+1/2*3+...+1/n-1/(n+1)=1-1/2+1/2-1/3+...+1/n-1/(n+1)  =...=1-1/(n+1)
1-1(n+1)=0,999=999/1000
(n+1-1)/(n+1)=999/1000=>n=999