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).

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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).

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))