求a1+a2+a3+......+a100的值
答案:3 悬赏:40 手机版
解决时间 2021-03-20 19:44
- 提问者网友：ミ烙印ゝ
- 2021-03-20 14:55
求a1+a2+a3+......+a100的值
最佳答案
- 五星知识达人网友：你哪知我潦倒为你
- 2021-03-20 15:41
a1+a2+a3+……+a100
=1/2×(1-1/3)+1/2×(1/3-1/5)+1/2×(1/5-1/7)+……+1/2×(199-1/201)
=1/2×(1-1/3+1/3-1/5+1/5-1/7+……+1/199-1/201)
=1/2×(1-1/201)
=1/2×200/201
=100/201
=1/2×(1-1/3)+1/2×(1/3-1/5)+1/2×(1/5-1/7)+……+1/2×(199-1/201)
=1/2×(1-1/3+1/3-1/5+1/5-1/7+……+1/199-1/201)
=1/2×(1-1/201)
=1/2×200/201
=100/201
全部回答
- 1楼网友：woshuo
- 2021-03-20 16:25
a1+a2+a3+......+a100
=
=(1/2 )*(200/201)
=100/201
- 2楼网友：持酒劝斜阳
- 2021-03-20 15:59
an=1/[(2n-1)(2n+1)]=(1/2)[1/(2n-1)-1/(2n+1)]
a1+a2+...+a100
=(1/2)[1/1-1/3+1/3-1/5+...+1/(2×100-1)-1/(2×100+1)]
=(1/2)(1- 1/201)
=100/201
a1+a2+...+a100
=(1/2)[1/1-1/3+1/3-1/5+...+1/(2×100-1)-1/(2×100+1)]
=(1/2)(1- 1/201)
=100/201
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