Question

(1/1+2)+(1/1+2+3)+(1/1+2+3+4)+(1/1+2+3+4....+n)

求(1/1+2)+(1/1+2+3)+(1/1+2+3+4)+(1/1+2+3+4....+n)的值

Answer 1

1:1
2:1+2
3:1+2+3
4:1+2+3+4
很明显，当第n个式子时：
n:1+2+3+4+···+n
=n（n+1)/2
1:1/1
2:1/（1+2)
3:1/（1+2+3)
4:1/ (1+2+3+4)
很明显，当第n个式子时：
n:1/(1+2+3+4+···+n)
=1/[n（n+1)/2]
=2/n(n+1)
(1/1+2)+(1/1+2+3)+(1/1+2+3+4)+(1/1+2+3+4....+n)
=2[1/(1*2)+1/(2*3)+……+1/n(n+1)] 用拆项法
=2[1-1/2+1/2-1/3+……+1/n-1/(n+1)]
=2*(1-1/(n+1)]
=2n/(n+1)

Answer 2

原题是不是：(1/1+2)+(1/1+2+3)+(1/1+2+3+4)+......(1/1+2+3+4......+n)
如果是就这样解:(1/1+2)+(1/1+2+3)+(1/1+2+3+4)+......(1/1+2+3+4....+n)
=2/(2*3)+2/(3*4)+2/(4*5.).....2/[n*(n+1)]
=2*[1/2-1/3+1/3-1/4+1/4-1/5.....1/n-1/*(n+1)]
=2*[1/2-1/*(n+1)]
=(n-1)/(n+1)

Answer 3

1 -2/(n+1)
当n趋于无穷时值为1
每一个单项的通式为：1/(1+2+3+4+···+n)=1/[n（n+1)/2]=2/(n(n+1))
1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+1/(1+2+3+4....+n)
=2[1/2-1/3+1/3-1/4+……+1/n-1/(n+1)]
=2*[1/2-1/(n+1) ]