中四符加數正(整數的二項定理)

1. 100! = 1*2*3*...*98*99*100
= 1*3*5*7*...*97*99*2*4*6*...*96*98*100
= 1*3*5*7*...*97*99* [2(1)][2(2)][2(3)]...[2(49)][2(50)]
= 1*3*5*7*...*97*99*2^50 (1*2*3*...*50)
=(2^50) (1 ‧3‧5‧‧‧99)‧50!)
2. p+[(p-2)p]/1!+[(p-4)p(p-1)]/2!+[(p-6)p(p-1)(p-2)]/3!+‧‧‧。
第k項 =[p-2(k-1)]p(p-1)(p-2)..(p-k+2)/(k-1)!
第(k+1)項 = (p-2k)p(p-1)(p-2)..(p-k+1)/(k)!
(b) 設S(n): p+[(p-2)p]/1!+[(p-4)p(p-1)]/2!+[(p-6)p(p-1)(p-2)]/3!+...+[p-2(n-1)]p(p-1)(p-2)..(p-n+2)/(n-1)! = [p(p-1)(p-2)‧‧‧(p-n+1)]/(n-1)!
當n = 1, LHS = p
RHS = p/0! = p
S(1) 成立
假設S(k) 成立, i.e. p+[(p-2)p]/1!+[(p-4)p(p-1)]/2!+[(p-6)p(p-1)(p-2)]/3!+...+[p-2(k-1)]p(p-1)(p-2)..(p-k+2)/(k-1)! = [p(p-1)(p-2)‧‧‧(p-k+1)]/(k-1)!
當n= k+1, LHS = p+[(p-2)p]/1!+[(p-4)p(p-1)]/2!+[(p-6)p(p-1)(p-2)]/3!+...+[p-2(k-1)]p(p-1)(p-2)..(p-k+2)/(k-1)! +(p-2k)p(p-1)(p-2)..(p-k+1)/(k)!
=[p(p-1)(p-2)‧‧‧(p-k+1)]/(k-1)! +(p-2k)p(p-1)(p-2)..(p-k+1)/(k)!
=[kp(p-1)(p-2)‧‧‧(p-k+1)]/k! + (p-2k)p(p-1)(p-2)..(p-k+1)/(k)!
=[(p)(p-1)(p-2)...(p-k+1)/k!] (k+p-2k)
= p(p-1)(p-2)...(p-k+1)(p-k)/k!
= p(p-1)(p-2)...(p-(k+1)+1)/[(k+1)-1]!
S(k+1)成立.
根據數學歸納法，S(n)成立，對所有正整數n.