求数列(2^2+1)/(2^2-1),(3^2+1)/(3^2-1),(4^2+1)/(4^2-1),…的前n项之和(要求用拆项法)

(n^2+1)/(n^2-1)
=1+2/(n^2-1)=1+(1/n-1 - 1/n+1)

从第n项入手，设第n项可拆为{an/(n-1)+b/(n+1)},整理得(an^2+an+bn-b)/(n^2-1).因为第n项为(n^2+1)/(n^2-1),故a=1,b=-1,通项可拆为n/(n-1)-1/(n+1),加和得，2/1-1/3+3/2-1/4+4/3-1/5+5/4-1/6+……+（n-2）/（n-3）-1/(-1)+(n-1)/(n-2)-1/n+n/(n-1)-1/(n+1),除开始两项和末尾两项有剩余，中间全部消去，得2+3/2-1/n-1/(n+1)