化簡 (1 question)

k(k+1)2 + (k+1)(3k+4)
= [ k(k+1) + (3k+4) ] (k+1)
= (k2 + k + 3k + 4) (k+1)
= (k2 + 4k + 4) (k+1)
= (k + 2)2 (k+1)

k(k+1)^2 + (k+1)(3k+4)
=k(k^2+2k+1)+3k^2+4k+3k+4
=k^3+2k^2+k+3k^2+4k+3k+4
=k^3+5k^2+7k+4

2007-11-01 20:48:16 補充：
(k 1)^2=k^2 2k 1係因為可以變成(k 1)(k 1)再變成k．ｋ＋ｋ＋ｋ＋１ =k^2 2k 1→係由此推出___

2007-11-01 20:49:03 補充：
打錯係(k 1)^2=k^2 2k 1係因為可以變成(k 1)(k 1)再變成k．ｋ＋ｋ＋ｋ＋１=k^2 2k 1→係由此推出___

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

k(k+1)2 + (k+1)(3k+4)
=3k2+(4k)(5)
=9k+20k
=29k

唔知係唔係咁呢

k(k+1)^2 + (k+1)(3k+4)
=k^3+2k^2+k+3k^2+7k+4
=k^3+5k^2+8k+4