(急)我想問呢條數學歸納法要點做?

當 n = 1 , 1(2)(3)(3+5) = 48 可被24整除。設 n = k 時 k(k+1)(k+2)(3k+5) = 24n 可被24整除 ,
當 n = k+1 ,
(k+1)(k+2)(k+3)(3(k+1)+5)
= (k+1)(k+2)(k+3)(3k+5 + 3)
= (k+3)(k+1)(k+2)(3k+5) + 3(k+1)(k+2)(k+3)
= k(k+1)(k+2)(3k+5) + 3(k+1)(k+2)(3k+5) + 3(k+1)(k+2)(k+3)
= k(k+1)(k+2)(3k+5) + 3[(k+1)(k+2)(3k+5) + (k+1)(k+2)(k+3)]
= 24n + 3(k+1)(k+2)(4k+8)
= 24n + 12(k+1)(k+2)(k+2)
由於(k+1)和 (k+2)中其中一個是偶數 , 可設 (k+1)(k+2) = 2m , 所以
= 24n + 12(2m)(k+2)
= 24n + 24m(k+2)
= 24(n + m(k+2)) 可被24整除。

由數學歸納法知對正整數 n ≥ 1 , n(n+1)(n+2)(3n+5)可被24整除。