以下是关于动态堆栈的摊销分析的文本片段.

If we are implementing stack as a dynamic array. Say that
inserting into the array costs 1, taking an element out of array costs
1, and the cost of resizing the array is the number of elements moved.
(Say that all other operations, like incrementing or decrementing
“top” are free). If we decide to double the size of the array when we
resize. Now, in any sequence of “n” operations, the total cost for
resizing is 1 +2 + 4+8 +…+(2^i) (i.e, 2 to power of i) for some 2^i
< n ( if all operations are pushes then (2 ^i) will be largest power
of 2 less than n). The sum is atmost 2n – 1. Adding in the additional
cost of “n” for inserting/removing, we get a total cost < 3n, and our
amoritzed cost per operations is < 3.

我的问题是作者如何得出结论,总和最多为2n -1.请求帮助举例或证明.

谢谢！

最佳答案 它是
geometric progression的总和：

SUM(q^k for k = 0 to n) = (q ^n+1 -1)/ (q-1)

那么在你的情况下你有：

SUM(2^k for k = 0 to i) = 2^(i+1) - 1 

然后,因为2 ^ i< ñ

2^(i+1) < 2n 

和

SUM(2^k for k = 0 to i) = 2^(i+1) - 1 < 2n - 1