Question

若公比为c的等比数列{an}的首项为a1=1,且满足an=[a(n-1)+a(n-2)]/2(n=3,4...)求数

若公比为c的等比数列{an}的首项为a1=1,且满足an=[a(n-1)+a(n-2)]/2(n=3,4...)求数列{nan}的前n项和

Answer 1

an=[a(n-1)+a(n-2)]/2
a(n-1)=a1q^(n-2)=q^(n-2)
an=q^(n-1)
a(n-2)=q^(n-3)
q^(n-1)=[q^(n-2)+q^(n-3)]/2
q²=(q+1)/2
2q²-q-1=0
1 -1
2 1
(q-1)(2q+1)=0
q=1 q=-1/2
1)q=1
an=a1=1
bn=nan=n
Tn=1+2+3+...+n=n+n(n-1)/2=(n²+n)/2
2)q=-1/2
an=(-1/2)^(n-1)
bn=nan=n(-1/2)^(n-1)
b1=1
b2=2(-1/2)^1
b3=3/(-1/2)^2
Tn= 1+2(-1/2)^1+3(-1/2)^2+..+n(-1/2)^(n-1)
-Tn/2= (-1/2) +2(-1/2)^2+...+(n-1)(-1/2)^(n-1)+n(-1/2)^n
3Tn/2=1+(-1/2)+(-1/2)^2+..+(-1/2)^(n-1) -n(-1/2)^n
=[1-(-1/2)^n]/(3/2)-n(-1/2)^n
Tn=4[1-(-1/2)^n]/9-2/3n(-1/2)^n

Answer 2

等比数列通项公式
An = c ^ (n-1)
2 * c ^ (n-1) = c ^ (n-2) + c ^ (n-3)
2 * c ^ 2 = c + 1
c = 1 或 -1/2
等比数列中公比不为1
所以 c = -1/2
设数列前n项和为Sn
Sn=A1+.A2+..+An
-1/2*Sn= +A2+...+An+An+1
3/2Sn = A1-An+1= 1 + (- 1/2)^(n)
Sn = 2/3 + (2/3) (- 1/2)^(n)