早教吧作业答案频道 -->数学-->
数列1/2,2/4,3/8,…,n/2^n,…的前n项和为?
题目详情
数列1/2,2/4,3/8,…,n/2^n,…的前n项和为?
▼优质解答
答案和解析
求数列{n/2^n}的前n项和
2^n表示2的n次方.
Sn = 1/2 + 2/2^2 + 3/2^3 + ...+ n/2^n
2Sn = 1 + 2/2^1 + 3/2^2 + ...+ n/2^(n-1)
所以 Sn
= 2Sn - Sn
= (1 + 2/2 + 3/2^2 + ...+ n/2^(n-1)) - (1/2 + 2/2^2 + 3/2^3 + ...+ n/2^n)
= 1 + 1/2 + 1/2^2 + ...+ 1/2^(n-1) - n/2^n
= 2 - 1/2^(n-1) - n/2^n
= 2 - (n+2)/2^n
2^n表示2的n次方.
Sn = 1/2 + 2/2^2 + 3/2^3 + ...+ n/2^n
2Sn = 1 + 2/2^1 + 3/2^2 + ...+ n/2^(n-1)
所以 Sn
= 2Sn - Sn
= (1 + 2/2 + 3/2^2 + ...+ n/2^(n-1)) - (1/2 + 2/2^2 + 3/2^3 + ...+ n/2^n)
= 1 + 1/2 + 1/2^2 + ...+ 1/2^(n-1) - n/2^n
= 2 - 1/2^(n-1) - n/2^n
= 2 - (n+2)/2^n
看了 数列1/2,2/4,3/8,...的网友还看了以下：