[ 本帖最後由 水龍蜥 於 16/9/2007 06:09 PM 編輯 ]作者: sapphire 時間: 16/9/2007 07:15 PM just expand it and u will get the ans
Assume P(k) is true i.e. k(k+1)(k+2)=3m for some +ve integer m
When n=k+1, (k+1)(k+2)(k+3) =k^3+6k^2+11k+6 =k^3+3k^2+2k+3k^2+9k+6 =k(k+1)(k+2)+3(k^2+3k+2) =3m+3(k+1)(k+2) (by the assumption) =3[m+(k+1)(k+2)] which is divisible by 3 becoz [m+(k+1)(k+2)] must be a +ve integer
By M.I., P(k) is true for n=1,2,3.....
[ 本帖最後由 sapphire 於 16/9/2007 07:22 PM 編輯 ]作者: 流映 時間: 26/9/2007 10:10 PM 「Assume P(k) is true」呢一句係唔出得街既...會俾人扣分作者: sapphire 時間: 26/9/2007 11:47 PM 用了四年也沒事,Marking也有這種字眼,會被扣分的話我早就死掉了作者: STEVE 時間: 27/9/2007 12:05 AM
