将原题写成：求和
\[ \sum_{n=2} ^{100} \sqrt{T_n} \]
其中
\[T_n = 1 + \frac{1}{n^2} + \frac{1}{(n + 1)^2}\]
通分相加
\[T_n = \frac{n^2(n+1)^2 + n^2 + (n+1)^2}{n^2(n+1)^2} = \frac{(n(n+1)+1)^2 }{n^2(n+1)^2} \]
这样，\[\sqrt{T_n} = 1 + \frac{1}{n(n + 1)} , \]
原式变成
\[99 + \frac{1}{2 \times 3}+ \frac{1}{3 \times 4}+ ... + \frac{1}{100 \times101} \]
\[ = 99 + \frac{1}{2} - \frac{1}{3}+ \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{100 }-\frac{1}{101}\]
\[ = 99 + \frac{1}{2} - \frac{1}{101}.\]
\[ \sum_{n=2} ^{100} \sqrt{T_n} \]
其中
\[T_n = 1 + \frac{1}{n^2} + \frac{1}{(n + 1)^2}\]
通分相加
\[T_n = \frac{n^2(n+1)^2 + n^2 + (n+1)^2}{n^2(n+1)^2} = \frac{(n(n+1)+1)^2 }{n^2(n+1)^2} \]
这样，\[\sqrt{T_n} = 1 + \frac{1}{n(n + 1)} , \]
原式变成
\[99 + \frac{1}{2 \times 3}+ \frac{1}{3 \times 4}+ ... + \frac{1}{100 \times101} \]
\[ = 99 + \frac{1}{2} - \frac{1}{3}+ \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{100 }-\frac{1}{101}\]
\[ = 99 + \frac{1}{2} - \frac{1}{101}.\]
