Question

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).

Answer 1

an=[(n+1)^2+1]/[(n+1)^2-1]=1+2/[(n+1)^2-1]=1+2/[n(n+2)]=1+(1/n)-(1/(n+2))
所以：
前n项之和=n+(1/1)-(1/3)+(1/3)-(1/5)+...+(1/n)-(1/(n+2))
=n+1-(1/(n+2))

追问

谢谢回答。计算是否有误，我把2代入，答案不对！

追答

哦，抱歉。应该是：
前n项之和=n+(1/1)-(1/3)+(1/2)-(1/4)+(1/3)-(1/5)+(1/4)-(1/6)...+(1/n)-(1/(n+2))

当n为偶数
前n项之和=n+[(1/1)-(1/3)+(1/3)-(1/5)+...+(1/(n-1))-(1/(n+1))]
+[(1/2)-(1/4)+(1/4)-(1/6)...+(1/n)-(1/(n+2))]
=n+1-(1/(n+1))+(1/2)-(1/(n+2))
=n+(3/2)-(1/(n+1))-(1/(n+2))

当n为奇数
前n项之和=前(n+1)项之和-a(n+1)
=(n+1)+(3/2)-(1/(n+2))-(1/(n+3))-[1+(1/(n+1))-(1/(n+3))]
=n+(3/2)-(1/(n+1))-(1/(n+2))

综合以上：
前n项之和=n+(3/2)-(1/(n+1))-(1/(n+2))