Sn=(2^2+1)/(2^2-1)+(3^2+1)/(3^2-1)+…+(n^2+1)/(n^2-1)+[(n+1)^

3个回答

  • 由最后一项化简整理得:

    [﹙n+1﹚²+1]/[﹙n+1﹚²-1]

    =[n²+2n+2]/[n﹙n+2﹚]

    =[n﹙n+2﹚+2]/[n﹙n+2﹚]

    =1+2/[n﹙n+2﹚]

    =1+[1/n-1/﹙n+2﹚]

    ∴以上每一项都可以拆成这种形式:

    Sn=[1+﹙1/1-1/3﹚]+[1+﹙1/2-1/4﹚]+[1+﹙1/3-1/5﹚]+[1+﹙1/4-1/6﹚]+……+﹛1+[1/﹙n-1﹚-1/﹙n+1﹚]﹜+﹛1+[1/n-1/﹙n+2﹚]﹜

    =1×n+﹛[1+1/2]+[-1/﹙n+1﹚-1/﹙n+2﹚]﹜

    =n+n/﹙n+1﹚+n/[2﹙n+2﹚]