✔ 最佳答案
1.
(a)用數學歸納法，證明對任意正整數n，
1^2*2+2^2*3+...+n^2(n+1)=n(n+1)(n+2)(3n+1)/12
Sol
當 n=1時
左=(1^2)*(1+1)=2
右=1*(1+1)*(1+2)*(3*1+1)/12=2
Son=1時為真
設n=k時為真即
1^2*2+2^2*3+...+k^2(k+1)=k(k+1)(k+2)(3k+1)/12
So
1^2*2+2^2*3+...+k^2(k+1)+ (k+1)^2(k+1+1)
= k(k+1)(k+2)(3k+1)/12+ (k+1)^2(k+1+1)
= k(k+1)(k+2)(3k+1)/12+ (k+1)^2(k+2)
=(k+1)(k+2)/12*[k(3k+1)+12(k+1)]
=(k+1)(k+2)/12*(3k^2+13k+12)
=(k+1)(k+2)/12*(k+3)(3k+4)
=(k+1)(k+2)(k+3)[3(k+1)+1]/12
Son=k+1時為真
(b) 推導出對任意正整數n，n(n+1)(n+2)(3n+1)可被24整除
Sol
當 n=1時
n(n+1)(n+2)(3n+1)=1*2*3*4=24
Son=1時為真
設n=k時為真即存在正整數p使得
k(k+1)(k+2)(3k+1)=24p
3k^4+10k^3+9k^2+2k=24p
3k^4=24p-10k^3-9k^2－2k
So
(k+1)(k+1+1)(k+2+1)[3(k+1)+1]
=(k+1)(k+2)(k+3)(3k+1+3)
=3k^4+22k^3+57k^2+62k+24
=24p-10k^3-9k^2－2k+22k^3+57k^2+62k+24
=24p+12k^3+60k+24
=24p+12(k^3+5k+2)
當k為奇數時k^3為奇數，5k為奇數，k^3+5k+2為2倍
當k為偶數時k^3為2倍數
So
存在整數q使得
(k+1)(k+1+1)(k+2+1)[3(k+1)+1]
=24p+12(k^3+5k+2)
=24p+24q
=24(p+q)
Son=k+1 時為真
