高一有關等比級數的問題~

Question

1．３＋２．４＋３．５＋...＋ｎ﹝ｎ＋２﹞＝ｎ﹝ｎ＋１﹞﹝ａｎ＋ｂ﹞/6，則數對﹝ａ，ｂ﹞＝？

為什麼我都看不懂呢~= =

Answer 1

原式=
[summation 2--> n+1] (n-1)(n+1)
= [summation 2--> n+1] (n^2) - [summation 2--> n+1] (1^2)
=[ (n+1)(n+2)(2n+3) / 6 - 1^2 ] - (n)
展開後得
(2n^3 + 9n^2 + 7n) / 6
即 n ( n + 1 )( n + 7 ) 故a=1 b=7

Answer 2

因 n(n+2)=n^2+2n
故 原式 1．３＋２．４＋３．５＋...＋ｎ﹝ｎ＋２﹞
= [1^2+2^2+3^2+..........+n^2] + 2[1+2+3+............+n]
= [n(n+1)(2n+1)] /6 + 2[n(n+1)/2]
= n(n+1)[(2n+1) /6+1]
= n(n+1)(2n+7) /6
故 a=2 b=7

Answer 3

設原式會=Σ(從1~n)k(k+1)
=Σk^2+2Σk
=n(n+1)(2n+1)/6+2n(n+1)/2
=n(n+1)(2n+7)
→ a=2 , b=7