24. 新考法 规律探究(10分)《庄子》一书中记载：“一尺之棰，日取其半，万世不竭.”其大意是：一尺长的木棍，每天截掉一半，永远也截不完.古人在两千多年前就有了数学极限思想，今天我们运用此数学思想研究下列问题.
(1)规律探索：如图1，正方形的边长为1，将它沿虚线剪掉一半，则$S_{阴影1}=1-\frac{1}{2}=\frac{1}{2}$；
如图2，在图1的基础上，将阴影部分再沿虚线剪掉一半，则$S_{阴影2}=1-\frac{1}{2}-(\frac{1}{2})^2=(\frac{1}{2})^2=\frac{1}{4}$；
依此类推，
如图3，$S_{阴影3}=1-\frac{1}{2}-(\frac{1}{2})^2-(\frac{1}{2})^3=\_\_\_\_\_\_$；
如图4，$S_{阴影4}=1-\frac{1}{2}-(\frac{1}{2})^2-(\frac{1}{2})^3-(\frac{1}{2})^4=\_\_\_\_\_\_$；
……
$S_{阴影n}=1-\frac{1}{2}-(\frac{1}{2})^2-(\frac{1}{2})^3-\dots-(\frac{1}{2})^n=\_\_\_\_\_\_$.

(2)规律应用：计算$\frac{1}{2}+(\frac{1}{2})^2+(\frac{1}{2})^3+\dots+(\frac{1}{2})^{10}$的值.

24. 解:(1)$S_{阴影3}=1-\frac{1}{2}-(\frac{1}{2})^2-(\frac{1}{2})^3=(\frac{1}{2})^3=\frac{1}{8}$；
$S_{阴影4}=1-\frac{1}{2}-(\frac{1}{2})^2-(\frac{1}{2})^3-(\frac{1}{2})^4=(\frac{1}{2})^4=\frac{1}{16}$；
$S_{阴影n}=1-\frac{1}{2}-(\frac{1}{2})^2-(\frac{1}{2})^3-\dots-(\frac{1}{2})^n=(\frac{1}{2})^n=\frac{1}{2^n}$.
故答案为$\frac{1}{8},\frac{1}{16},\frac{1}{2^n}$.
(2)因为$1-\frac{1}{2}-(\frac{1}{2})^2-(\frac{1}{2})^3-\dots-(\frac{1}{2})^{10}=\frac{1}{2^{10}}$，所以$\frac{1}{2}+(\frac{1}{2})^2+(\frac{1}{2})^3+\dots+(\frac{1}{2})^{10}=1-\frac{1}{2^{10}}$.
解题大招
利用数形结合,将算式转化为计算图形的面积,从而解决问题.