pomoc

[Gość: gosc]

mam pytanie jak rozłożyć na sumę ulamkow iloraz 1/n(n+1)(n+2)?

[Gość: joa]

1/n(n+1)(n+2)=a/n +b/(n+1)+c/(n+2) <=>a(n+1)(n+2) +b(n(n+2)+
+cn(n+1)=1<=>(a+b+c)n^2+(3a+2b+c)n+2a <=> a+b+c=0 i 3a+2b+c=0 i 2a=1<=>a=1/2,
b=-1, c=-1/2 .
1/n(n+1)(n+2)=(1/2)/n + (-1)/(n+1)+ (1/2)/(n+2)