裂项就行了1/(x²-a²)=1/2a[1/(x-a)-1/(x+a)]而[1/(x-a)](n)=∑(-1)^n·n!/(x-a)^(n+1)[1/(x+a)](n)=∑(-1)^n·n!/(x+a)^(n+1)故y(n)=∑(-1)^n·n!/2a[1/(x-a)^(n+1)+1/(x+a)^(n+1)]