如何算这个阶乘的求和∑

Question

如题所述

Answer 1

∑(i:1->81) (160-i)!/(81-i)!
=∑(i:1->81) (160-i)(159-i)...(82-i)
=∑(i:1->81) (82-i)(83-i)....(160-i)
=(1/80) ∑(i:1->81) [ (82-i)(83-i)....(160-i)(161-i) - (81-i)(82-i)(83-i)....(160-i) ]
=(1/80) [ (82-1)(83-1)....(160-1)(161-1) - (81-81)(82-81)(83-81)....(160-81) ]
=(1/80) (81)(82)....(160)
=(1/80) [160!/80!]