:数列累加法求An=An-1+1/(n-1)(n+1) (n>2)的通项公式

an=a(n-1)+1/[(n-1)(n+1)]=a(n-1)+(1/2)[1/(n-1)-1/(n+1)]
2an-2a(n-1)=1/(n-1)-1/(n+1)
2a(n-1)-2a(n-2)=1/(n-2)-1/n
…………
2a2-2a1=1/1-1/3
累加
2an-2a1=1/1-1/3+1/2-1/4+1/3-1/5+...+1/(n-2)-1/n+1/(n-1)-1/(n+1)
2an-2a1=[1/1+1/2+1/3+...+1/(n-1)]-[1/3+1/4+1/5+...+1/(n-1)+1/n+1/(n+1)]
2an-2a1=1+1/2-1/n-1/(n+1)
=3/2-(2n+1)/[n(n+1)]
=[3n(n+1)-(4n+2)]/[2n(n+1)]
=(3n²-n-2)/[2n(n+1)]
=(n-1)(3n+2)/[2n(n+1)]
an=a1+(n-1)(3n+2)/[4n(n+1)]
这就是所求的通项公式,如果已知a1,就可以求出确定的an