Verified answer

when n=k+1

右方=(k+1)(k+2)^2............(*)

左方=1．4+2．7+3．10+．．．．．+ｋ(3ｋ＋1)+(k+1)(3k+4)............(**)

......=k(k+1)^2+(k+1)(3k+4)............(***)

......=(k+1)(k(k+1)+3k+4)

......=(k+1)(k^2+4k+4)

......=(k+1)(k+2)^2

P(k+1) is true if P(k) is true for any positive integer k

by principle of MI ,P(n) is true for all positive integers n

(*)在P(n)條式中右方代ｎ＝ｋ＋１

(**)在P(n)條式中左方代ｎ＝ｋ＋１時，由於佢只係加到n項,咁你就要幫佢加多個n+1項

記得要轉寫ｋ,但尼步唔一定要寫，如果係明，可以直接寫下步

(***)因為你已經假設左P(k) is true,所以你就將P(k)個result代入去

2007-09-15 19:16:18 補充：

k(k 1)^2 (k 1)(3k 4)let k 1=xkx^2 x(3k 4)x(kx 3k 4)抽通項

當n = k+1,

左方=1．4＋2．7＋3．10＋．．．．．＋ｋ（ 3ｋ＋1）+ (k+1)[3(k+1)+1]

＝ｋ（ｋ＋1）＾2 + (k+1)(3k+4)

=(k+1)(k^2+k+3k+4)

=(k+1)(k^2+4k+4)

=(k+1)(k+2)^2

=(k+1)[(k+1)+1]^2

所以p(k+1)成立

如下:

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