1/(3+6)+1/(3+6+9)+1/(3+6+9+12)+...+1/(3+6+9+.+100)
答案:2 悬赏:50 手机版
解决时间 2021-01-14 03:42
- 提问者网友:藍了天白赴美
- 2021-01-13 18:27
1/(3+6)+1/(3+6+9)+1/(3+6+9+12)+...+1/(3+6+9+.+100)
最佳答案
- 五星知识达人网友:西风乍起
- 2021-01-13 19:58
裂项法:
1/(3+6)+1/(3+6+9)+1/(3+6+9+12)+...+1/(3+6+9+.+102)
=2/3*(1/6+1/12+1/20+……+1/34*35)
=2/3*(1/2-1/3+1/3-1/4+1/4-1/5+……+1/34-1/35)
=2/3*(1/2-1/35)
=2/3*33/70
=11/35
1/(3+6)+1/(3+6+9)+1/(3+6+9+12)+...+1/(3+6+9+.+102)
=2/3*(1/6+1/12+1/20+……+1/34*35)
=2/3*(1/2-1/3+1/3-1/4+1/4-1/5+……+1/34-1/35)
=2/3*(1/2-1/35)
=2/3*33/70
=11/35
全部回答
- 1楼网友:十鸦
- 2021-01-13 20:17
考察一般项第k项:
1/(3+6+...+3k)=1/[3(1+2+...+k)]=1/[3k(k+1)/2]=(2/3)/[k(k+1)]=(2/3)[1/k -1/(k+1)]
1/3+1/(3+6)+...+1/(3+6+...+3n)
=(2/3)[1/1-1/2+1/2-1/3+...+1/n -1/(n+1)]
=(2/3)[1- 1/(n+1)]
=2n/[3(n+1)]
你题目没写完,如果原题是求到1/(3+6+...+3n),那么就结束了,如果是求极限,那么:
1/3+1/(3+6)+...+1/(3+6+...+3n)
=(2/3)[1- 1/(n+1)]
=(2/3)- (2/3)/(n+1)
n->+∞ n+1->+∞ 2/3为定值,(2/3)/(n+1)->0
(2/3)- (2/3)/(n+1)->2/3
lim1/3+1/(3+6)+...+1/(3+6+...+3n)=2/3
n->+∞
希望对你能有所帮助。追问我是6年级的学生
1/(3+6+...+3k)=1/[3(1+2+...+k)]=1/[3k(k+1)/2]=(2/3)/[k(k+1)]=(2/3)[1/k -1/(k+1)]
1/3+1/(3+6)+...+1/(3+6+...+3n)
=(2/3)[1/1-1/2+1/2-1/3+...+1/n -1/(n+1)]
=(2/3)[1- 1/(n+1)]
=2n/[3(n+1)]
你题目没写完,如果原题是求到1/(3+6+...+3n),那么就结束了,如果是求极限,那么:
1/3+1/(3+6)+...+1/(3+6+...+3n)
=(2/3)[1- 1/(n+1)]
=(2/3)- (2/3)/(n+1)
n->+∞ n+1->+∞ 2/3为定值,(2/3)/(n+1)->0
(2/3)- (2/3)/(n+1)->2/3
lim1/3+1/(3+6)+...+1/(3+6+...+3n)=2/3
n->+∞
希望对你能有所帮助。追问我是6年级的学生
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