1/(1*3*5)+1/(3*5*7)+1/(5*7*9)+ …+1/(93*95*97)+1/(95*97*99)＝多少
答案:3 悬赏:40 手机版
解决时间 2021-02-22 07:13
- 提问者网友：感性作祟
- 2021-02-21 18:16
1/(1*3*5)+1/(3*5*7)+1/(5*7*9)+ …+1/(93*95*97)+1/(95*97*99)＝多少
最佳答案
- 五星知识达人网友：独行浪子会拥风
- 2021-02-21 19:06
1/(1*3*5)+1/(3*5*7)+1/(5*7*9)+ …+1/(93*95*97)+1/(95*97*99)
= (1/4)*[1/(1*3) - 1/(3*5)] + (1/4)*[1/(3*5) - 1/(5*7)] + ... + (1/4)*[1/(95*97) - 1/(97*99)]
= (1/4)*[1/(1*3) - 1/(97*99)]
= 800/9603
= (1/4)*[1/(1*3) - 1/(3*5)] + (1/4)*[1/(3*5) - 1/(5*7)] + ... + (1/4)*[1/(95*97) - 1/(97*99)]
= (1/4)*[1/(1*3) - 1/(97*99)]
= 800/9603
全部回答
- 1楼网友：枭雄戏美人
- 2021-02-21 20:59
1/(1×3×5)+1/(3×5×7)+1/(5×7×9)+...+1/(95×97×99)
=1/4×[(1/1×3-1/3×5)+(1/3×5-1/5×7)+...+(1/95×97-1/97×99)]
=1/4×(1/1×3-1/97×99)
= 800/9603
=1/4×[(1/1×3-1/3×5)+(1/3×5-1/5×7)+...+(1/95×97-1/97×99)]
=1/4×(1/1×3-1/97×99)
= 800/9603
- 2楼网友：山河有幸埋战骨
- 2021-02-21 19:38
∵1/[(n－2)·n·(n＋2)]＝(1/8)·[1/(n－2)－2/n＋1/(n＋2)]
∴
1/(1·3·5)＝(1/8)·(1/1－2/3＋1/5)
＝(1/8)·(1/1－1/3－1/39＋1/43)
＝557/6708
∴
1/(1·3·5)＝(1/8)·(1/1－2/3＋1/5)
＝(1/8)·(1/1－1/3－1/39＋1/43)
＝557/6708
参考资料：刚刚搞错了是 800/9603
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