计算:4/(1*3*5)+4/(3*5*7)+4/(5*7*9)+…4/(93*95*97)+4/(95*97*99)
答案:3 悬赏:40 手机版
解决时间 2021-10-13 22:37
- 提问者网友:精神病院里
- 2021-10-13 03:09
计算:4/(1*3*5)+4/(3*5*7)+4/(5*7*9)+…4/(93*95*97)+4/(95*97*99)
最佳答案
- 五星知识达人网友:三千妖杀
- 2021-05-07 23:25
请注意到:
1/[n(n+2)(n+5)]
=(1/4){[(n+4)-n]/[n(n+2)(n+4)]}
=(1/4){1/[n(n+2)]-1/[(n+2)(n+4)]}。
于是:
原式
=[1/(1×3)-1/(3×5)]+[1/(3×5)-1/(5×7)]+···+[1/(95×67)-1/(97×99)]
=1/(1×3)-1/(97×99)
=1/3-1/[97×(100-1)]
=1/3-1/(9700-97)
=1/3-1/9603
=(3201-1)/9603
=3200/9603。
1/[n(n+2)(n+5)]
=(1/4){[(n+4)-n]/[n(n+2)(n+4)]}
=(1/4){1/[n(n+2)]-1/[(n+2)(n+4)]}。
于是:
原式
=[1/(1×3)-1/(3×5)]+[1/(3×5)-1/(5×7)]+···+[1/(95×67)-1/(97×99)]
=1/(1×3)-1/(97×99)
=1/3-1/[97×(100-1)]
=1/3-1/(9700-97)
=1/3-1/9603
=(3201-1)/9603
=3200/9603。
全部回答
- 1楼网友:轮獄道
- 2018-12-01 07:19
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
- 2楼网友:你哪知我潦倒为你
- 2020-11-25 13:45
原式第n项={1/[n(n+2)]-1/[(n+2)(n+4)]}。
n=1___n代入原式,化简,销项得原式=1/(1×3)-1/(97×99)=3200/9603
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