方楠法公式
方楠法公式：[k/(a(a+b))]+[k/((a+b)(a+2b))]+[k/((a+2b)(a+3b))]+……+[k/((a+NB)(a+(n+1)b))]
=((n+1)k)/(a(a+(n+1)b)
公式證明
[k/(a(a+b))]+[k/((a+b)(a+2b))]+[k/((a+2b)(a+3b))]+……+[k/((a+nb)(a+(n+1)b))]
=(k/b)*{[1/a]-[1/(a+b)]+[1/(a+b)]-[1/(a+2b)]+……+[1/(a+nb)]-[1/(a+(n+1)b)]}
=(k/b)*{[(n+1)b]/[a(a+(n+1)b]}
=((n+1)k)/(a(a+(n+1)b)
運用方楠法做題
1、[1/(1*2)]+[1/(2*3)]+……+[1/(100*101)]=100/(1*101)=100/101
2、[8/(1*5)]+[8/(5*9)]+[8/(9*13)+]……+[8/(401*405)]=(8*101)/(1*405)=808/405
