求数列{(n+1)^2+1⼀(n+1)^2-1}的前n项和Sn

等式可以化成(n^2+2n+2)/(n^2+2n)=1+(1/n-1/(n+2))
Sn=n+(1-1/3+1/3-1/5…………+1/n-1/(n+2))
=n+(1-1/(n+2))=n+(n+1)/(n+2)
解答完毕，请采纳

先分离系数，再裂项
an=[(n+1)^2+1]/[(n+1)^2-1]
={[(n+1)²-1]+2}/[(n+1)²-1]
=1+2/[(n+1)²-1]
=1+2/[n(n+2)]
=1+[1/n-1/(n+2)]
Sn=a1+a2+........+an
=(1+1/1-1/3)+(1+1/2-1/4)+(1+1/3-1/5)+........+[1+1/(n-1)-1/(n+1)]+[1+1/n-1/(n+2)]
=n+[1-1/3+1/2-1/4+1/3-1/5+......+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=n+1+1/2-1/(n+1)-1/(n+2)
=3/2+n-(2n+3)/[(n+1)(n+2)]

不明白再追问吧，是不容易，很高兴能帮助你

请问最后一个1是在分母上 还是单独的-1

如果是在分母上的话就可以裂项了 如果是单独的-1 就是变调和数列求和 貌似超过高中范围了

对不起，我是霸刀封天
Sn=n+[n/(n+1)+n/2(n+2)]
写错了，抱歉
哎，对不住了