35.【难度：★★★★☆】探究与发现：
观察下列等式的规律，解答下列问题：
$a_1=\frac{1}{2}×(\frac{2}{1}+\frac{2}{2});a_2=\frac{1}{2}×(\frac{2}{2}+\frac{2}{3});a_3=\frac{1}{2}×(\frac{2}{3}+\frac{2}{4});$
$a_4=\frac{1}{2}×(\frac{2}{4}+\frac{2}{5});a_5=\frac{1}{2}×(\frac{2}{5}+\frac{2}{6});···$
(1)第6个等式$a_6=$，第100个等式$a_{100}=$；
(2)第n个等式$a_n=$；(用含n的代数式表示)
(3)设$S_1=a_1 - a_2,S_2=a_3 - a_4,S_3=a_5 - a_6,···,S_{1011}=a_{2021} - a_{2022}$，求$S_1 + S_2 + S_3 + ··· + S_{1011}$的值.

35. 解:(1)$\frac{1}{2}×(\frac{2}{6}+\frac{2}{7})$ $\frac{1}{2}×(\frac{2}{100}+\frac{2}{101})$
(2)$\frac{1}{2}×(\frac{2}{n}+\frac{2}{n+1})$
(3)由(2),得$a_n=\frac{1}{2}(\frac{2}{n}+\frac{2}{n+1})=\frac{1}{n}+\frac{1}{n+1}$,所以$S_1=a_1-a_2=(1+\frac{1}{2})-(\frac{1}{2}+\frac{1}{3})=1-\frac{1}{3}$,$S_2=a_3-a_4=(\frac{1}{3}+\frac{1}{4})-(\frac{1}{4}+\frac{1}{5})=\frac{1}{3}-\frac{1}{5}$,$S_3=a_5-a_6=(\frac{1}{5}+\frac{1}{6})-(\frac{1}{6}+\frac{1}{7})=\frac{1}{5}-\frac{1}{7}$,$\dots S_{1011}=a_{2021}-a_{2022}=(\frac{1}{2\ 021}+\frac{1}{2\ 022})-(\frac{1}{2\ 022}+\frac{1}{2\ 023})=\frac{1}{2\ 021}-\frac{1}{2\ 023}$.所以$S_1+S_2+S_3+\dots+S_{1011}=(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+\dots+(\frac{1}{2\ 021}-\frac{1}{2\ 023})=1-\frac{1}{2\ 023}=\frac{2\ 022}{2\ 023}$.