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

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