[急!幫手]一條數學問題

Let An=an^3+bn^2+cn+d
A1=a+b+c+d=4 ...(1)
A2=8a+4b+2c+d=10 ...(2)
A3=27a+9b+3c+d=18...(3)
A4=64a+16b+4c+d=28...(4)

(2)-(1)=> 7a+3b+c=6 ...(5)
(3)-(2)=> 19a+5b+c=8 ...(6)
(4)-(3)=> 37a+7b+c=10 ...(7)

(6)-(5)=> 12a+2b=2 ...(8)
(7)-(6)=> 18a+2b=2 ...(9)
(9)-(8)=>6a=0 or a=0

Putting into (8), b=1
Putting into (5), c=3
Putting into (1), d=0

So An=n^2+3n
A99999=99999^2+3*99999=10,000,099,998

2009-06-21 23:13:04 補充：
2 , 3 , 5 , 6 , 10 , 11 , 17 , 18 , 26 , 27 ,..
求第4000項
An=n+1+[integer part of (n-1)/2]^2
A4000=1999^2+4000+1=4,000,002

(99999)(100002)=10000099998

The nth term is n(n + 3). So the 99999term is 99999(99999 + 3)
= (99999)(100002).

200000


設 n=項 nX2+2


　　so 99999X2+2=200000