1/1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+61)=?

来源：百度知道编辑：UC知道时间：2024/05/12 04:10:53

要写出步骤！！！

1/1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+61)
= 2/1*2 + 2/2*3 + 2/3*4 + ... + 2/61*62
= (2/1 - 2/2) + (2/2 - 2/3) + (2/3 - 2/4)+...+(2/61-2/62)
= 2/1 - 2/62
= 61/31

1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3...+n)
=1+1*2/(2*3)+1*2/(3*4)+...+1*2/[n*(1+n)]
=2[1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/n-1/(n+1)]
=2[1/2+1/2-1/(n+1)]
=2-2/(n+1)
=2n/(n+1)

将n=61带入可得
61/31

1/( )+1/( )+1/( )+1/( )+1/( )+1/( )=1 (1/2005-1)(1/2004-1)........(1/3-1)(1/2-1) (1-1/100)(1-1/99)(1-1/98)......(1-1/3 1/8=1/( )+1/( )1 1/10=1/( )+1/( ）1/12=1/( )+1/( )+1/( ) (1-1/2)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000) (1/2003-1)(1/2002-1)(1/2001-1)...*(1/1001-1)(1/1000-1) (1-1/4)(1-1/9)(1-1/16)^(1-1/4008004)(1-1/4012009) 1+1/2+1+1/3+1+1/4+......+1/100=? 1/1+1/2+1/3+1/4+。。。。+1/N 是多少 1-1/4 -1/8-1/16-1/32-1/64