[n+1)^2+1] /[(n+1)^2-1]

原式=[1+2/(2^2-1)]+[1+3/(3^2-1)]+....+[1+2/(n^2-1)]+[1+2/((n+1)^2-1)]
=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)]
=n+3/2-1/(n+1)-1/(n+2)

1 /[(n+1)^2-1]
化为1/n(n+2) --平方差公式得到
化为0.5[(1/n)-1/(n+2)]
通项可化为[（n+1)^2+0.5[(1/n)-1/(n+2)]
即2^2+3^2_4^2+...+(n+1)^2
+0.5[1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7+...+1/(n-3)-1/(n-2)+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
2^2+3^2_4^2+...+(n+1)^2 这个有简便公式，老师应该讲过吧？
0.5[1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7+...+1/(n-3)-1/(n-2)+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2)]销去一些项后，
得 1/2+1/3-[1/（n+1)]-[1/(n+2)]
两部分相加

d