There are two ways to make the step. One is to recognize this as a geometric series, and know the rule:
sum=(1-rn)/(1-r)
The other is to have played around enough with powers of two to know that if you add up a bunch of them starting from 1, you get the next one, minus one.
是序列Mk=.....的第N项 而这个第 N 项本身就是第一项为 1(2^0) 和公比=2 的几何级数之和。 这个 sum(Mn) 是
=a[(r^n)-1]/[r-1]
其中 a 是第一项和 r 的公比
=1*[(2^n)-1]/[2-1]
Mn=2^n - 1
Actually,
Mn=2^0+2^1+.........+2^(n-1)+2^(n-2)
is the Nth term of the sequence Mk=..... And this Nth term itself is a sum of a geometrical progression whose 1st term is 1(2^0) and common ratio=2. And this sum(Mn) is
发布评论
评论（4）
步骤来自
mn =2n−1 +2n−2 +...+22 +2+1。
m
n = 2n − 1
有两种方法可以实现该步骤。一是认识到这是一个几何级数，并知道规则：
sum=(1-rn)/(1-r)
另一个是玩够了 2 的幂来知道如果你从 1 开始将一堆数字相加，你就会得到下一个，减去一。
The step is from
mn =2n−1 +2n−2 +...+22 +2+1.
to
mn = 2n − 1
There are two ways to make the step. One is to recognize this as a geometric series, and know the rule:
sum=(1-rn)/(1-r)
The other is to have played around enough with powers of two to know that if you add up a bunch of them starting from 1, you get the next one, minus one.
几何级数的前 n 项之和有一个公式。
There is a formula for the sum of the first n terms of a geometric series.
这只是人们多年来弄清楚的级数关系之一：
你可以把它想象成二进制数的和：
It's just one of the relations of series that people have figured out over the years:
You can think of it a lot like the sum of binary numbers:
实际上，
是序列
Mk=.....的第N项而这个第 N 项本身就是第一项为 1(2^0) 和公比=2 的几何级数之和。
这个
sum(Mn)是其中 a 是第一项和 r 的公比
Actually,
is the
Nthterm of the sequenceMk=.....And this
Nthterm itself is a sum of a geometrical progression whose 1st term is1(2^0)andcommon ratio=2.And this
sum(Mn)iswhere a is 1st term and r common ratio