Question

Fie numarul a=1/(1*2)+1/(2*3)+......+1/n(n+1)
N>1
a) sa se determine n€n astfel incat {a}=0,999

Answer 1

1/(1*2)=1/1-1/2
1/(2*3)=1/2-1/3
1/(3*4)=1/3-1/4
...
1/(n(n+1))=1/n-1/(n+1)

Deci a=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/n-1/(n+1)).

Se reduc: -1/2 cu +1/2; -1/3 cu +1/3; -1/4 cu +1/4; ...; -1/n cu 1/n.
Obtinem deci a=1/1-1/(n+1)=1-1/(n+1).

0<=a<1, oricare ar fi n numar natural, deci a={a}=0,999.

a=0,999 <=> 1-1/(n+1)=0,999 <=> 0,001=1/(n+1) <=> n+1=1/0,001 <=> n+1=1000 <=> n=999.

In concluzie, n=999.