Question

an为A.P,Sn为前n项和,a1=2,当n属于自然数时,3Sn+1=1+2Sn,则an=?Sn=?

Answer 1

S(2)=a1+a2=2+a2=(1+2S1)/3=(1+2a1)/3=5/3

所以:a2=5/3-a1=-1/3

当n>1时:

3S(n+1)=1+2Sn ---(1)

S(n+1)=Sn+a(n+1) ---(2)

(1)-(2)*2得:

S(n+1)=1-2*a(n+1)

所以S(n)=1-2*a(n)

a(n+1)=S(n+1)-S(n)=1-2*a(n+1)-[1-2*a(n)]=2*[a(n)-a(n+1)]

得:

a(n+1)=(2/3)*a(n)

当n>1时,此数列为等比数列:公比:(2/3)

因此通项式为:an=a2*(2/3)^(n-2)=-3/4*(2/3)^n [n>1]

即:an=-3/4*(2/3)^n [n>1时],或 an=2 [n=1时]

Sn=1+3/2*(2/3)^n