S(N)=1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+.+1/[1+2+.+(n+1)]
S(n)=(1+2+.+(n+1)=[1+(n+1)]*n/2=[(n+2)*(n+1)]/2
1/[S(n)]=2/[(n+2)*(n+1)]
=2*{1/[(n+2)*(n+1)]}
=2*[1/(n+1)-1/(n+2)]
代入“S(N)=1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+.+1/[1+2+.+(n+1)]”中得：
S(N)=2*[1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+.+1/(n+1)-1/(n+2)]
=2*[1/2-1/(n+2)]
=1-2*[1/(n+2)]
=1-2/(n+2)額、不太看得懂也、*全是次方還是有的地方表示乘號?S(n)算的是分母相加？直接分之一、、得到的是什么？不好意思哦、我比較笨、規(guī)律1：二個相鄰的整數(shù)的導(dǎo)數(shù)之差例1：1/3-1/4=4/12-3/12 =(4-3)/12 =1/12即：1/(3*4)規(guī)律2：一個連續(xù)的倒數(shù)之和例2：1/6+1/12+1/20+1/30+1/42=1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7=1/2-1/7