1/2+1/6+1/12+1/20+1/30+1/42的求和公式是怎么推导的?为什么是N/N+1?

原式=(1/1-1/2)+(1/2-1/3)……[1/(N-1)-1/N]+[1/N-1/(N+1)]
=1-1/(N+1)=N/N+1

1/2+1/6+1/12+1/20+1/30+1/42的求和公式是怎么推导的？
只题用到的是裂项相消
原式＝1/(1*2)+1/(2*3)+1/(3*4)+……＋1/(6*7)
因为1/(n*(n+1))=1/n-1/(n+1)
然后代入就可以发现原式＝1/1-1/2+1/2-1/3……＋1/6－1/7
抵消后只剩下1-1/7＝6/7
为什么是N/N+1？剩下的基本上就是1-1/(1+n)=n/(1+n)

1/2+1/6+1/12+1/20+1/30+1/42
=1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+1/(6*7)
=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+(1/5-1/6)+(1/6-1/7)
=1-1/7
=6/7

原式＝1/(1*2)+1/(2*3)+1/(3*4)+……＋1/(6*7)
因为1/(n*(n+1))=1/n-1/(n+1)
然后代入就可以发现原式＝1/1-1/2+1/2-1/3……＋1/6－1/7
抵消后只剩下1-1/7＝6/7