用拆项法求数列(2^2+1)⼀(2^2-1),(3^2+1)⼀(3^2-1),(4^2+1)⼀(4^2-1),…的前n项之和。 麻烦过程！谢谢！

a(n) = [(n+1)^2 + 1]/[(n+1)^2 - 1] = 1 + 2/[(n+1)^2 - 1]
=1 + 2/[(n+2)n]
=1 + 1/n - 1/(n+2),

s(n) = a(1)+a(2)+a(3)+a(4)+...+a(n-3)+a(n-2)+a(n-1)+a(n)
=n + [1/1-1/3 + 1/2-1/4 + 1/3-1/5 + 1/4-1/6 + ... + 1/(n-3)-1/(n-1) + 1/(n-2)-1/n + 1/(n-1)-1/(n+1) + 1/n-1/(n+2)]
=n + 1/1 + 1/2 - 1/(n+1) - 1/(n+2)
=n + 3/2 - 1/(n+1) - 1/(n+2)

通项为(n²+1)/(n²-1)=1+2/(n-1)(n+1)=1+2[1/(n-1)-1/(n+1)]
所以前n项和为n+2[1-1/3)+(1/2-1/4)+(1/3-1/5)+...+1/(n-1)-1/(n+1)]
=n+2[(3/2-1/n-1/(n+1)].

不明白追问，明天回答，休息了，晚安.