【题文】用
![](http://img.shitiku.com.cn/uploads/allimg/20200324/20200324192506-24925.png)
表示自然数
![](http://img.shitiku.com.cn/uploads/allimg/20200324/20200324192506-90383.png)
的所有因数中最大的那个奇数,例如:9的因数有1,3,9,
![](http://img.shitiku.com.cn/uploads/allimg/20200324/20200324192506-72583.png)
,10的因数有1,2,5,10,
![](http://img.shitiku.com.cn/uploads/allimg/20200324/20200324192506-96471.png)
,那么
![](http://img.shitiku.com.cn/uploads/allimg/20200324/20200324192507-97227.png)
;
.
【解析】此题答案为:85,
![](http://img.shitiku.com.cn/uploads/allimg/20200324/20200324192507-62163.png)
(4
n-1)
据题中对g(n)的定义,判断出g(n)=g(2n),且若n为奇数则g(n)=n,利用等差数列的前n项和公式及逐差累加的方法及等比数列的前n项和公式求出g(1)+g(2)+g(3)+…+g(2
n-1),令n=4求出g(1)+g(2)+g(3)+…+g(15).
解:由g(n)的定义易知g(n)=g(2n),且若n为奇数则g(n)=n
令f(n)=g(1)+g(2)+g(3)+…g(2
n-1)
则f(n+1)=g(1)+g(2)+g(3)+…g(2
n+1-1)=1+3+…+(2
n+1-1)+g(2)+g(4)+…+g(2
n+1-2)
=2
n[1+(2
n+1-1)]/2+g(1)+g(2)+…+g(2
n+1-2)=4
n+f(n)
即f(n+1)-f(n)=4
n分别取n为1,2,…,n并累加得f(n+1)-f(1)=4+4
2+…+4
n=
![](http://img.shitiku.com.cn/uploads/allimg/20200324/20200324192508-78173.png)
=
![](http://img.shitiku.com.cn/uploads/allimg/20200324/20200324192507-62163.png)
(4
n-1)
又f(1)=g(1)=1,所以f(n+1)=
![](http://img.shitiku.com.cn/uploads/allimg/20200324/20200324192507-62163.png)
(4
n-1)+1
所以f(n)=g(1)+g(2)+g(3)+…g(2
n-1)=
![](http://img.shitiku.com.cn/uploads/allimg/20200324/20200324192507-62163.png)
(4
n-1-1)+1
令n=4得
g(1)+g(2)+g(3)+…+g(15)=
![](http://img.shitiku.com.cn/uploads/allimg/20200324/20200324192507-62163.png)
(4
3-1)+1=85
故答案为85,
![](http://img.shitiku.com.cn/uploads/allimg/20200324/20200324192507-62163.png)
(4
n-1).