求数列1/2,3/4,5/8,7/16……2n-1/2^n前n项和
人气:498 ℃ 时间:2020-03-10 04:27:42
解答
an=(2n-1)/2^nSn=1/2+3/4+5/8+...+(2n-3)/2^(n-1)+(2n-1)/2^n1/2Sn= 1/4+3/8+...+(2n-3)/2^n+(2n-1)/2^(n+1)上式减下式:Sn-1/2Sn=1/2+2/4+2/8+2/16+...+ 2/2^n-(2n-1)/2^(n+1)=1/2-(2n-1)/2^(n+1)-1+2(1/2^1+1/2^2+1...
推荐
- 数列求和:1/2+3/4+5/8+7/16+.+(2n-1)/(2^n)
- 求数列1/2,3/4,5/8,7/16……2n-1/2^n,……的前n项和Sn
- 数列1又1/2,2又1/4,3又1/8,4又1/16,.的前n项和为多少?
- 数列1×(1/2),2×(1/4),3×(1/8),4×(1/16)```````的前N项和为?
- 数列1×1/2,2×1/4,3×1/8,4×1/16求前N项和.
- 请问system two-state least squares怎么翻译?
- 就在那一瞬间为题写一篇作文!
- 以What I want to speak out most 写一篇120字英语短文
猜你喜欢