數列問題 x2 數列問題 x2 20點

1.
a2-a1=17-13=4
a3-a2=23-17=6
a4-a3=31-23=8
a5-a4=41-31=8
so
a2-a1=2*2
a3-a2=2*3
a4-a3=2*4
a5-a4=2*5
so a[n+1]-an=2*(n+1)=2n+2
since a1-a0=2 so a0=11
set an=pn^2+qn+r
a0=r=11
a1=p+q+r=13
a2=4p+2q+r=17
so p=1,q=1,r=11
so an=n^2+n+11

2.
a1+a2=12,a2+a3=-6
a1+a1r=12,a1r+a1r^2=-6
so (a1r+a1r^2)/(a1+a1r)=r=-6/12=-1/2
so lim(Sn)=24/(1-(-1/2))=16

圖片參考：http://i40.tinypic.com/29azlhc.jpg

13 = 13 = 13 + [2*4+2(1-2)](1-1)/2
17 = 13 + 4 = 13 + [2*4+2(2-2)](2-1)/2
23 = 13 + 4 + 6 = 13 + [2*4+2(3-2)](3-1)/2
31 = 13 + 4 + 6 + 8 = 13 + [2*4+2(4-2)](4-1)/2
41 = 13 + 4 + 6 + 8 + 10 = 13 + [2*4+2(5-2)](5-1)/2
所以第n項 = 13 + [2*4+2(n-2)](n-1)/2
= 13 + (2n+4)(n-1)/2
= 13 + (2n2 + 4n - 2n - 4)/2
= 13 + n2 + n - 2
= n2 + n + 11

圖片參考：http://i44.tinypic.com/n1eafk.jpg

設公比是R,則
a1 + a1*R = 12.....(1)
a1*R + a1*R2 = - 6......(2)
(2)/(1):
R = -6/12 = -1/2,代入(1)得
a1 = 24
所以該數列的無窮項之和 = 24 / [1- (-1/2)] = 16

http://i40.tinypic.com/29azlhc.jpg:n

http://i44.tinypic.com/n1eafk.jpg:mum